Question

0.110 M in HBr and 0.100 M in HCHO2 Express your answer to three decimal places. pH = .

. Part B 0.155 M in HNO2 and 9.0×10−2 M in HNO3 Express your answer to two decimal places. pH = .

Part C 0.175 M in HCHO2 and 0.220 M in HC2H3O2 Express your answer to two decimal places. pH =

Part D 5.5×10−2 M in acetic acid and 5.5×10−2 M in hydrocyanic acid Express your answer to two decimal places. pH =

Answer 1

Part A

Since HBr is a strong acid. So, it will dissociate fully and [H⁺] = 0.110 M

But HCOOH is a weak acid and it dissociates as:

HCOOH H⁺ + HCOO^-

I(M) 0.100 0.110

C(M) . -x +x +x

E(M) 0.100-x 0.110+x x

K_a = [H⁺][ HCOO^-] / [HCOOH]

1.8 *10^-4 = (0.110 +x)x /(0.100-x)

But Ka is very small, so

1.8 *10^-4 = 0.11 x/ 0.1

1.8 * 10^-5 = 0.11 x

x = 16.36 * 10^-5

Total [H⁺] = 0.11 + 0.00016 = 0.11016 M

pH = -log[H⁺] = -log (0.11016) = 0.958

2.

Since HNO₃ is a strong acid. So, it will dissociate fully and [H⁺] = 0.09 M

But HCOOH is a weak acid and it dissociates as:

HCOOH H⁺ + HCOO^-

I(M) 0.155 0.090

C(M) . -x +x +x

E(M) 0.155-x 0.09+x x

K_a = [H⁺][ NO₂^-] / [HNO₂]

4.5 * 10^-4 = (0.09 +x)x /(0.155-x)

But Ka is very small, so

4.5 *10^-4 = 0.09 x/ 0.155

7.0 * 10^-5 = 0.09 x

x = 7.75* 10^-5

Total [H⁺] = 0.09 + 0.000075 = 0.0900775 M

pH = -log[H⁺] = -log (0.0900775) = 1.045

3. Both are weak acids.

HCOOH dissociates as:

HCOOH H⁺ + HCOO^-

I(M) 0.175 0 0

C(M) . -x +x +x

E(M) 0.175-x x x

K_a = [H⁺][ HCOO^-] / [HCOOH]

1.8 *10^-4 = (x)x /(0.175-x)

But Ka is very small, so

1.8 *10^-4 = x² / 0.175

3.15* 10^-5 = x²

x = 0.00561

CH₃COOH dissociates as

CH₃COOH H⁺ + cH₃COO^-

I(M) 0.220 0 0

C(M) -x +x +x

E(M) 0.220-x x x

Ka =[H⁺][cH₃COO^-] / [CH₃COOH]

1.8 *10^-5 = x² / (0.220-x)

Since Ka is very small. So,

1.8 *10^-5 = x² / (0.220)

3.96 * 10^-6 = x²

x = 0.00199

Total [H⁺] = 0.00561 + 0.00199 = 0.0075999

pH = -log (0.0075999) = 2.12

4. Since both are weak

So, Ka(CH₃COOH) = 1.8 * 10^-5 = x² / (0.055-x)

1.8 * 10^-5 = x² / (0.055)

9.9 * 10^-7 = x²

x = 0.000995 = [H⁺]

Similarly, Ka(HCN) = 6.2 * 10^-10 = x² / (0.055)

3.41 * 10^-11 = x²

x= 0.00000584

total [H⁺] =0.000995 + 0.00000584 = 0.001

pH = 3.00