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

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 =

Replacing:

6,17x10⁻⁵ =

Solving:

[H⁺] = 1,543x10⁻⁴

As pH = -log [H⁺]

pH = 3,81

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