Question

1.Calculate the pH of a buffer solution that is 0.249 M in HCN and 0.175 M in KCN. For HCN, Ka = 4.9×10⁻¹⁰ (pKa = 9.31).

2.Calculate the pH of a buffer solution that is 0.210 M in HC2H3O2 and 0.200 M in NaC2H3O2. (Ka for HC2H3O2 is 1.8×10−5.)

3.Consider a buffer solution that is 0.50 M in NH3and 0.20 M in NH4Cl. For ammonia, pKb=4.75. Calculate the pH of 1.0 L of the original buffer, upon addition of 0.020 mol of solid NaOH.

4.A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.065 mol of NaOH to the original buffer.

5.Calculate the pH of the buffer that results from mixing 56.3 mL of a 0.292 M solution of HCHO2 and 20.0 mL of a 0.734 M solution of NaCHO2. The Ka value for HCHO2 is 1.8×10−4.

6.Consider a buffer solution that is 0.50 M in NH3and 0.20 M in NH4Cl. For ammonia, pKb=4.75. Calculate the pH of 1.0 L of the solution, upon addition of 10.00 mL of 1.0 MHCl.

7.A 1.0-L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. Calculate the pH of the solution upon addition of 78.9 mL of 1.00 MHCl to the original buffer.

Answer 1

Question 1.

Calculate the pH of a buffer solution that is 0.249 M in HCN and 0.175 M in KCN.

For HCN, Ka = 4.9×10−10 (pKa = 9.31)

Note that this appears to be a buffer;

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Now, use the Henderson Hasselbach equation

pH = pKa + log(KCN/HCN)

pKa = 9.31

substitute all data:

pH = 9.31 + log(0.175/0.249)

pH = 9.1568