Chem help please ::: Calculate the pH of a buffer that is 0.125 M in NaHCO3 and 0.105 in Na2CO3.?

please show me the steps and how you got to the answer thank you so much!!!!



also.... 2) A beaker with 155 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 0.1M . A student adds 8.90 ml of a 0.340 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.



10 points to best answer!!!! thanks

even if you only help with one that is a great help!!!Chem help please ::: Calculate the pH of a buffer that is 0.125 M in NaHCO3 and 0.105 in Na2CO3.?
(HCO3)-1 --%26gt; H+ %26amp; (CO3)-2

[0.125M] ....... [H+] .....[0.105M]



Ka for (HCO3)-1 from the back of my book: 5.6 e-11



5.6e-11 = [H+] [0.105] / [0.125]



[H+] = 6.67e-11



pH+ = 1.0176



your answer = 1.02 , (only 2 sig figs in the Ka)

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also.... 2)

using ';The total molarity of acid and conjugate base in this buffer is 0.1M'; %26amp; ';buffer with a pH of 5.00';



H C2H3O2 --%26gt; H+ %26amp; (C2H3O2)-1

...[0.1 - X] .........[1.0e-5] [X]



using the ';pKa = 4.76'; gives the Ka for acetic acid = 1.74 e-5

which is close to what most people use as the Ka



1.74 e-5 = [1.0e-5] [X] / [0.1 - X]



(1.74 e-5) (0.1 - X) = [1.0e-5] [X]



1.74e-6 - 1.74e-5 X = 1.0e-5 X



1.76e-6 = 2.74e-5 X



[C2H3O2] = X = 0.0642 Molar



%26amp; so [H C2H3O2] = [0.10 - X] = 0.0358 Molar

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...........H C2H3O2 --%26gt; H+ %26amp; (C2H3O2)-1

initial [0.0358M] ........[H+]......[0.0642M] @ 0.155 litres

gives

0.005549 moles........[H+]......0.009951 moles

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but when you add 0.00890 Litres of 0.340mol/Litre HCl, then youi are adding 0.003026 moles of H+'s....

which shifts the reaction to the left, increasing the HC2H3O2 by 0.003026 moles, %26amp; decreasing the (C2H3O2)-1 by the same 0.003026 moles

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...................H C2H3O2 --%26gt; H+ %26amp; (C2H3O2)-1

initial : 0.005549 mols.....[H+]......0.009951 mols

change:+0.003026 mols.................-0.003026mols

______________________________________?br>
final eq : 0.008575 mol .......[H+] .....0.006925 mols



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I am going to intentially make a mistake that will still give the same right answer. I won't divide the numerator's 0.006925 mols by the new total volume, %26amp; I won't divide the denominator's 0.008575 mol by the new total volume to get the new molarities. since the work cancels itself out... So:



1.74 e-5 = [H+] [0.006925] / [0.008575]



[H+] = 2.15 e-5



pH+ = 4.6666



Your answer is possibly: pH+ = 4.67

(I am assuming that the buffer had more sig figs than ';0.1M';)

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but, interpret the phrase: ';How much will the pH change? ';



might mean that your answer should be phrased:

By adding the 8.90 ml of a 0.340 M HCl, the buffer changes from a pH+ = 5.00 to a pH+ of 4.67.... ie. ';it decreases in pH+ by 0.03 pH+ units';