Here's the problem I am given: A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.
The main thing I'm having a problem with is figuring out how to find the moles of acid and base using the henderson hasselbalch equation. I found the ratio of conjugate base to conjugate acid (1.74) but that's as far as I've gotten. If someone could show me how to do that I would really appreciate it.
Thanks for your help!Problem Help! Henderson Hasselbalch equation.
Ok I would change thew molarity as it wree to mmoles of ...in other words the 120 mL X 0.1 M = 12 mmoles total for the two forms acid and base. Since the pH is 5 then the amount of each form can be determined by th following H/H 5.0 = 4.76 + log base /12 - base since the pH is merely a function of the ratio of mmoles since the two forms are in the same volume. so 5.0 - 4.76 = log base /12-base
so 1.74 = base /12-base
20.88 = 2.74 base base = 7.62 mmoles and acid = 12 - 7.62 = 4.38 moles
check
pH = 4.76 + log 7.62/4.38 = 4.76 + log 1.739 = 4.76 + 0.24 = 5
so now the pH after adding the HCl will be driven more acidic because the ratio uis going to change 6.6mL x 0.3 M = 1.98 mmole of H+ so adding this to the buffer will convert 1.98 moles of the base form ( acetate ) to the acid form so the pH will now be
pH = 4.76 + log[ 7.62-1.98]/4.38 + 1.98 mmoles = .76 + log 5.64/6.36
= 4.76 -0.052 = 4.708 it changes by - .29 pH units
Wednesday, September 21, 2011
How to get rid of weeds?
I have a patch of weeds that I need to kill. I am going to put grass there so I don't want to change the pH of the soil. Is there a quick way to do this? How to get rid of weeds?
everyone is right about needing to pull them out. the easiest way to do that is to thoroughly soak the base of the weed(s) by putting a hose on it (barely turned on) for about 10 minutes, and in many cases you can just grab the weed at the base and pull it out, root and all. use a weed knife to make sure you get the entire root. don't shake the weed once you get it out, throw the entire thing away, dirt and all, to ensure you don't drop any seeds. weed killer usually puts the root into hibernation mode, and it will thrive once again with enough watering. How to get rid of weeds?
You could dig for their roots. That way, the weeds can't grow back. I suggest a hoe if it grew to an amazing height.
The safest and most efficient way is to take a shovel and dig them up by roots and all and shake the dirt back out. Some people prefer chemicals but they are bad for your soil and the bees. If it's a larger area use a rototiller and then rake out as much of the weeds as you can and throw them in a compost pile elsewhere.
The quickest way, of course, is to till the weeds under with a roto-tiller, but if you were wondering about weed killer, you could use that too without changing the pH of the soil. You have to prep the soil for the new lawn seed anyway, so...remove the big weeds and till the rest under. The new lawn should choke out the weeds once it is established.
I use an herbicide called 2-4d it won't affect grass,ponds,wildlife,or house pets but kills weeds quick.
Pour boiling water on them. They will be brown the next day, then dig them up and make sure you get all the roots.
Good Luck!
If the weeds are green %26amp; activly growing us shuld use ';Roundup'; - very effective %26amp; breaks down quickly.rodents acrylic nails gel nails
everyone is right about needing to pull them out. the easiest way to do that is to thoroughly soak the base of the weed(s) by putting a hose on it (barely turned on) for about 10 minutes, and in many cases you can just grab the weed at the base and pull it out, root and all. use a weed knife to make sure you get the entire root. don't shake the weed once you get it out, throw the entire thing away, dirt and all, to ensure you don't drop any seeds. weed killer usually puts the root into hibernation mode, and it will thrive once again with enough watering. How to get rid of weeds?
You could dig for their roots. That way, the weeds can't grow back. I suggest a hoe if it grew to an amazing height.
The safest and most efficient way is to take a shovel and dig them up by roots and all and shake the dirt back out. Some people prefer chemicals but they are bad for your soil and the bees. If it's a larger area use a rototiller and then rake out as much of the weeds as you can and throw them in a compost pile elsewhere.
The quickest way, of course, is to till the weeds under with a roto-tiller, but if you were wondering about weed killer, you could use that too without changing the pH of the soil. You have to prep the soil for the new lawn seed anyway, so...remove the big weeds and till the rest under. The new lawn should choke out the weeds once it is established.
I use an herbicide called 2-4d it won't affect grass,ponds,wildlife,or house pets but kills weeds quick.
Pour boiling water on them. They will be brown the next day, then dig them up and make sure you get all the roots.
Good Luck!
If the weeds are green %26amp; activly growing us shuld use ';Roundup'; - very effective %26amp; breaks down quickly.
Please explain what is in a buffer...?
and discuss the function of a buffer.
How will pH change when small amounts of acids or bases are added to the buffer solution?Please explain what is in a buffer...?
A buffer is a solution of a weak acid and it's conjugate weak base. The pH of the buffer is such that it is within one pH unit of the pKa of the acid (aka the weak acid and weak base forms are present in near-equal amounts). This means that when you add small amounts of acid or base to the buffer, the weak acid/base will sort of ';soak up'; the extra H+ (or OH-) ions and the result will be little to no change in the pH of the solution.
So the purpose of a buffer is to make a solution where the pH stays stable even when small amounts of acid or base are added.
How will pH change when small amounts of acids or bases are added to the buffer solution?Please explain what is in a buffer...?
A buffer is a solution of a weak acid and it's conjugate weak base. The pH of the buffer is such that it is within one pH unit of the pKa of the acid (aka the weak acid and weak base forms are present in near-equal amounts). This means that when you add small amounts of acid or base to the buffer, the weak acid/base will sort of ';soak up'; the extra H+ (or OH-) ions and the result will be little to no change in the pH of the solution.
So the purpose of a buffer is to make a solution where the pH stays stable even when small amounts of acid or base are added.
Adding a Strong Acid to a Buffer?
have the solution, but I still dont understand some of the steps.
A beaker with 115mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M . A student adds 4.30mL of a 0.490M HCl solution to the beaker. How much will the pH change? The Pk_a of acetic acid is 4.76.
SOLUTION:
pH = pKa log acetate / acid
5.00 = 4.76 log acetate / acid
acetate / acid =10^0.24= 1.74
acetate acid = 0.1
Solve the system
acetate = 0.0365 M
acid = 0.0635 M
*HOW DID THEY FIND THAT IT'S .0365 M %26amp; .0635 M???
*CAN SOMEONE SHOW ME HOW TO THE MATH FOR IT?????
moles acetate = .115 L x 0.0365 M = 0.00420 mol
moles acid = .115 L x 0.0635 M = 0.00730 mol
Moles H added = .00430 L x 0.490 M = 0.00211mol
CH3COO- H --%26gt; CH3COOH
moles acetate = 0.00420 - 0.00211 = 0.00209 mol
moles acid = 0.00730 0.00211 = 0.00941 mol
*WHY SUBTRACT FOR ACETATE BUT THEN ADD FOR ACID??
Total volume = 115 4.30 = 119.3 mL = 0.1193 L
concentration acetate = 0.00209/0.1193 = 0.0175 M
concentration acid = 0.00941 / 0.1193 = 0.0789 M
pH = 4.76 log 0.0175 / 0.0789 = 4.11Adding a Strong Acid to a Buffer?
[CH3COO-] / [CH3COOH] = 1.74
[CH3COOH] + [CH3COO-] = 0.1
[CH3COO-] = 0.1 - [CH3COOH]
We put this value in the 1st equation :
0.1 -[CH3COOH] / [CH3COOH] = 1.74
we multuply the left and the right side by [CH3COOH]
0.1 - [CH3COOH] = 1.74 [CH3COOH]
0.1 = 2.74 [CH3COOH]
[CH3COOH] = 0.0365 M
[CH3COO-] + 0.0365 = 0.1
[CH3COO-] = 0.1 - 0.0365 = 0.0635 M
the effect of the added 0.00211 mol of H+ would be to decrease the moles of CH3COO- by 0.00211 and increase the moles of CH3COOH by 0.00211 by the reaction :
CH3COO- + H+ %26gt;%26gt; CH3COOH
A beaker with 115mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M . A student adds 4.30mL of a 0.490M HCl solution to the beaker. How much will the pH change? The Pk_a of acetic acid is 4.76.
SOLUTION:
pH = pKa log acetate / acid
5.00 = 4.76 log acetate / acid
acetate / acid =10^0.24= 1.74
acetate acid = 0.1
Solve the system
acetate = 0.0365 M
acid = 0.0635 M
*HOW DID THEY FIND THAT IT'S .0365 M %26amp; .0635 M???
*CAN SOMEONE SHOW ME HOW TO THE MATH FOR IT?????
moles acetate = .115 L x 0.0365 M = 0.00420 mol
moles acid = .115 L x 0.0635 M = 0.00730 mol
Moles H added = .00430 L x 0.490 M = 0.00211mol
CH3COO- H --%26gt; CH3COOH
moles acetate = 0.00420 - 0.00211 = 0.00209 mol
moles acid = 0.00730 0.00211 = 0.00941 mol
*WHY SUBTRACT FOR ACETATE BUT THEN ADD FOR ACID??
Total volume = 115 4.30 = 119.3 mL = 0.1193 L
concentration acetate = 0.00209/0.1193 = 0.0175 M
concentration acid = 0.00941 / 0.1193 = 0.0789 M
pH = 4.76 log 0.0175 / 0.0789 = 4.11Adding a Strong Acid to a Buffer?
[CH3COO-] / [CH3COOH] = 1.74
[CH3COOH] + [CH3COO-] = 0.1
[CH3COO-] = 0.1 - [CH3COOH]
We put this value in the 1st equation :
0.1 -[CH3COOH] / [CH3COOH] = 1.74
we multuply the left and the right side by [CH3COOH]
0.1 - [CH3COOH] = 1.74 [CH3COOH]
0.1 = 2.74 [CH3COOH]
[CH3COOH] = 0.0365 M
[CH3COO-] + 0.0365 = 0.1
[CH3COO-] = 0.1 - 0.0365 = 0.0635 M
the effect of the added 0.00211 mol of H+ would be to decrease the moles of CH3COO- by 0.00211 and increase the moles of CH3COOH by 0.00211 by the reaction :
CH3COO- + H+ %26gt;%26gt; CH3COOH
Adding a strong acid to buffer?
A beaker with 200 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.20 mL of a 0.260 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
please help??? I'm lost....Adding a strong acid to buffer?
5.00 = 4.760 + log [CH3COO-] / [CH3COOH]
10^0.24 = 1.74 = [CH3COO-] / [CH3COOH]
[CH3COOH] + [CH3COO-] = 0.100
[CH3COOH] = 0.100 - [CH3COO-]
1.74 = [CH3COO-] / 0.100 - [CH3COO-]
0.174 - 1.74 [CH3COO-] = [CH3COO-]
[CH3COO-] = 0.0635 M
[CH3COOH]= 0.100 - 0.0635 =0.0365 M
moles acetate = 0.200 L x 0.0635 =0.0127
moles acetic acid = 0.200 x 0.0365 =0.00730
moles HCl = 0.00620 L x 0.260 M=0.00161
CH3COO- + HCl %26gt;%26gt; CH3COOH
moles acetate = 0.0127 - 0.00161 =0.0111
moles acetic acid = 0.00730 + 0.00161 =0.00891
total volume =0.206 L
cocnentration acetate = 0.0111 / 0.206 =0.0538 M
concentration acetic acid = 0.00891 / 0.206 =0.0433 M
pH = 4.760 + log 0.0538 / 0.0433 =4.85
please help??? I'm lost....Adding a strong acid to buffer?
5.00 = 4.760 + log [CH3COO-] / [CH3COOH]
10^0.24 = 1.74 = [CH3COO-] / [CH3COOH]
[CH3COOH] + [CH3COO-] = 0.100
[CH3COOH] = 0.100 - [CH3COO-]
1.74 = [CH3COO-] / 0.100 - [CH3COO-]
0.174 - 1.74 [CH3COO-] = [CH3COO-]
[CH3COO-] = 0.0635 M
[CH3COOH]= 0.100 - 0.0635 =0.0365 M
moles acetate = 0.200 L x 0.0635 =0.0127
moles acetic acid = 0.200 x 0.0365 =0.00730
moles HCl = 0.00620 L x 0.260 M=0.00161
CH3COO- + HCl %26gt;%26gt; CH3COOH
moles acetate = 0.0127 - 0.00161 =0.0111
moles acetic acid = 0.00730 + 0.00161 =0.00891
total volume =0.206 L
cocnentration acetate = 0.0111 / 0.206 =0.0538 M
concentration acetic acid = 0.00891 / 0.206 =0.0433 M
pH = 4.760 + log 0.0538 / 0.0433 =4.85
Suppose a mixture's hydrogen ion concentration is increased by a factor of 100...?
...By How much and in what direction will the pH change?
its to the left right? but is it by 100? or am i missing something?
(Ok, my last algebra question for the night, I've been spending hours on this, now it's almost 2 AM and I need to know what is going on so I can get some sleep and go to class at 8 AM. Can anybody help me out here? It would be SO appreciated. Thanks)Suppose a mixture's hydrogen ion concentration is increased by a factor of 100...?
The pH will change by two. It will go down. If you start out at pH 5 you will end up at pH3.
pH is defined as the negative log ouf the hydrogen ion concentration. If it changes by a factor of 100, the log changes by two. If the concentration goes up, the pH will go down.Suppose a mixture's hydrogen ion concentration is increased by a factor of 100...?
think you mean by a factor of 10 not 100.
pH is log (base 10) hydrogen ion concentration
log 100 is 2
so pH is altered by 2 and more acid so reduced by 2 (to the lefT)
its to the left right? but is it by 100? or am i missing something?
(Ok, my last algebra question for the night, I've been spending hours on this, now it's almost 2 AM and I need to know what is going on so I can get some sleep and go to class at 8 AM. Can anybody help me out here? It would be SO appreciated. Thanks)Suppose a mixture's hydrogen ion concentration is increased by a factor of 100...?
The pH will change by two. It will go down. If you start out at pH 5 you will end up at pH3.
pH is defined as the negative log ouf the hydrogen ion concentration. If it changes by a factor of 100, the log changes by two. If the concentration goes up, the pH will go down.Suppose a mixture's hydrogen ion concentration is increased by a factor of 100...?
think you mean by a factor of 10 not 100.
pH is log (base 10) hydrogen ion concentration
log 100 is 2
so pH is altered by 2 and more acid so reduced by 2 (to the lefT)
Chemistry question involving buffers?
A beaker with 175mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100M . A student adds 6.80mL of a 0.430M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Chemistry question involving buffers?
[CH3COOH] + [CH3COO-]= 0.100 M
pH = pKa + log [CH3COO-] / [CH3COOH]
5.00 = 4.760 + log [CH3COO-] / [CH3COOH]
10^0.24 =1.74 = [CH3COO-]/ [CH3COOH]
solving this system
[CH3COO-] = 0.0635 M
[CH3COOH]= 0.0365 M
Moles acetate = 0.0635 M x 0.175 L=0.0111
moles acetic acid = 0.0365 M x 0.175 L= 0.00639
moles HCl = 0.00680 L x 0.430 M=0.00292
CH3COO- + H+ %26gt;%26gt; CH3COOH
moles acetate = 0.0111 - 0.00292 =0.00818
moles acetic acid = 0.00639 + 0.00292 =0.00931
total volume = 175 + 6.80=181.8 mL = 0.1818 L
[acetate]= 0.00818 / 0.1818 = 0.0500 M
[acetic acid ]= 0.00931 / 0.1818 = 0.0512 M
pH = 4.760 + log 0.0500/ 0.0512 = 4.750
[CH3COOH] + [CH3COO-]= 0.100 M
pH = pKa + log [CH3COO-] / [CH3COOH]
5.00 = 4.760 + log [CH3COO-] / [CH3COOH]
10^0.24 =1.74 = [CH3COO-]/ [CH3COOH]
solving this system
[CH3COO-] = 0.0635 M
[CH3COOH]= 0.0365 M
Moles acetate = 0.0635 M x 0.175 L=0.0111
moles acetic acid = 0.0365 M x 0.175 L= 0.00639
moles HCl = 0.00680 L x 0.430 M=0.00292
CH3COO- + H+ %26gt;%26gt; CH3COOH
moles acetate = 0.0111 - 0.00292 =0.00818
moles acetic acid = 0.00639 + 0.00292 =0.00931
total volume = 175 + 6.80=181.8 mL = 0.1818 L
[acetate]= 0.00818 / 0.1818 = 0.0500 M
[acetic acid ]= 0.00931 / 0.1818 = 0.0512 M
pH = 4.760 + log 0.0500/ 0.0512 = 4.750
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