I have the answers: 21) 10.2 22) pH changes from 2.73 to 3.76, concentration of
ID: 694749 • Letter: I
Question
I have the answers: 21) 10.2 22) pH changes from 2.73 to 3.76, concentration of CH3COOH goes up from 0.198M to 0.1998M, concentration of CH3COO- goes down from 0.00188M to 1.745x10^-4M. *** I need to know how to do these with explainations and how to put it in the calculator. I need a break down of how to do these Thank you! 21. Calculate the pH of 0.15 M solution of Na,HPO,if Ka(H.PO?) = 6.3x 10 22. An acid-base equilibrium system is created by dissolving 0.20mol CH COOH in water and diluting the resulting solution to a volume of 1.0 L. What is the effect of adding 0.020 mol CH,COO (aq) to this solution How will pH change (calculate pH before and after the addition. Ka of CH,COOH is 1.76 x10)? How will concentrations of CH,COOH and CH,COO at equilibrium change?Explanation / Answer
21) pka of H2PO4^- = -log(6.3*10^-8) = 7.2
pH of Na2HPO4 = 7+1/2(pka +logC)
= 7+1/2(7.2+log0.15)
= 10.2
22) concentration of CH3COOH = 0.2/1 = 0.2 M
pka of CH3COOH = -logka
= -log(1.76*10^-5)
= 4.75
pH = 1/2(pka-logC)
= 1/2(4.75-log0.2)
= 2.74
after addition
pH = pka + log(CH3COO-/CH3COOH)
= 4.75+log(0.02/0.2)
= 3.75
pH changes from 2.74 to 3.75.
before addition,
CH3COOH <----> CH3COO- + H3O+
initial 0.2 0 M 0 M
change - x + x + x
equil 0.2 - x x x
ka = [CH3COO-][H3O+]/[CH3COOH]
(1.76*10^-5) = x*x/(0.2-x)
x = 0.00187
[CH3COO-] = x = 0.00187
[ch3cooh] = 0.2-x = 0.2- 0.00187 = 0.198 M
after addition,
(1.76*10^-5) = (0.02+x)*x/(0.2-x)
x = 0.000174
[CH3COO-] = 0.02+x = 0.02+0.000174 = 0.020174 M
[ch3cooh] = 0.2-0.000174 = 0.199 M