b. Does the concentration of water change significantly when enough acetic acid

Question

b. Does the concentration of water change significantly when enough acetic acid is added to water to make a 0.1 M solution of acetic acid? No 5. What is the pH of a 0.1 M aqueous solution of HC H.0, which has a K -1.78 x 10.? 6. What is the pH of a 0.1 M aqueous solution of NaC,H.0.? 7. Calculate the pH of an aqueous solution containing 0.10 M HC.H.0, and 0.10 M NaC.H.O Phosphoric acid is a triprotic acid. It has three separate values of K. What does triprotic mean and what reactions relate to each of the K values? 8.

Explanation / Answer

5.

HC2H3O2= 0.1M

K= 1.78x10^-5

for weak acids

[H+]= square root og KxC

[H+] = square root of (1.78x10^-5x0.1)

[H+] =1.334x10^-3M

-log[H+] = -log(1.334x10^-3)

PH= 2.87

6)

NaC2H3O2 =0.1M

NaC2H3O2 ----------------------- Na+ + C2H3O2-

C2H3O2- + H2O-------------------- HC2H3O2 + OH-

0.1                                             0                0

-x                                               +x                +x

0.1-x                                          +x +x

Ka= 1.78x10^-5

KaxKb= Kw    where Kw = ionic product of water = 1.0x10^-14

kb = kw/ka= 1.0x10^-14/1.78x10^-5

kb= 5.62x10^-10

Kb = [HC2H3O2 ][OH-]/[C2H3O2 -]

5.62x10^-10= x*x/(0.1-x)

for solving the equation

x=7.5x10^-6

[OH-] = 7.5x10^-6M

-log[OH-]= -log(7.5x10^-6)

POH=5.12

PH+POH= 14

PH= 14-POH

PH= 14- 5.12

PH= 8.88

7)

HC2H3O2 = 0.1M

NaC2H3O2= 0.1M

Ka= 1.78x10^-5

-log(ka) = -log(1,78x10^-5)

Pka= 4.75

PH= PKa + log[salt]/[acid]

PH= 4.75+log(0.1/0.1)

PH= 4.75.

.

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