Calculate the CHANGE in PH for the following:?

Calculate the change in PH that occurs:



50 ml of water is added to 20 ml of a 0.50 mol/L solution of hydrochloric acid. Calculate the CHANGE in ph that occurs.



Please show the steps how to do this problem.Calculate the CHANGE in PH for the following:?
this question was actually asked by someone else a while ago:-) nways...









use the formula;



c1V1 = c2V2



c = concentration (molL^-1)

V = volume (L)

1 = before addition of water

2 = after addition of water



0.50molL^-1 x (20/1000)L = c2 x [(20 + 50)/1000]L

0.01mol = c2 x 0.07L

c2 = 0.01mol / 0.07L

c2 = 0.0007 molL^-1

or 7 x 10^-4 molL^-1



pH = -log[H^+]



pH(before) = -log[0.50] = 0.301029996

pH(after) = -log[7 x 10^-4] = 3.15490196



thus, change in pH =%26gt; pH(after) - pH(before)

= 3.15490196 - 0.301029996

= 2.853871964

thus, the answer is 2.85 (3s.f.)



hop this helps:-)







Hope this helps:-)Calculate the CHANGE in PH for the following:?
pH of 0.50 M HCl = -log(0.5) = 0.30

after adding 50 mL of water the Molarity is 0.50 x (20/70) = 0.143.

pH of 0.143 M HCl = -log(0.143) = 0.85

change in pH = +0.55



20 mL x 0.50 M = 10 millimoles of HCl

add 50 mL to get 70 mL

70 mL x ? M = 10 millimoles HCl (still the same amount of HCl, only more dilute)

? M = 10 millimoles / 70 mL = 0.143 M

microsoft word

hip hop abs program