Question

If 1.2 mL of 0.01 M HCl is added to a 350 mL solution containing 0.500 M succinic acid and 0.200 M potassium succinate, what is the resulting pH of the solution? (Identify the acid and the conjugate base, write out all equations). Ka succinic acid= 6.17 x 10^-5

Answer 1

Answer:

pH is 3,81

Explanation:

The equilibrium:

Succinic acid ⇄ Succinate + H⁺; ka = 6,17x10⁻⁵

The initial moles of each compound are:

Succinic acid: 0,350L × 0,500M = 0,175 moles

Succinate: 0,350L × 0,200M = 0,07 moles

The addition of HCl will consume succinate and produce succinic acid, thus:

Moles of HCl:

0,0012L × 0,01M = 1,2x10⁻⁵ moles

Succinic acid: 0,175 moles +  1,2x10⁻⁵ moles = 0,175012 moles

Succinate: 0,07 moles -  1,2x10⁻⁵ moles = 0,069988 moles

The equilibrium equation is:

ka = $\frac{[succinate][H^{+}]}{[Succinic Acid]}$

Replacing:

6,17x10⁻⁵ = $\frac{[0,069988][H^{+}]}{[0,175012]}$

Solving:

[H⁺] = 1,543x10⁻⁴

As pH = -log [H⁺]

pH = 3,81

I hope it helps!