Question

A buffer solution contains 0.496 M KHCO3 and 0.340 M K2CO3. If 0.0585 moles of potassium hydroxide are added to 250. mL of this buffer, what is the pH of the resulting solution

Answer 1

Answer:

pH = 10.9

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:

$n_{CO_3^{2-}}=0.34mol/L*0.250L+0.0585mol=0.1435mol\\\\n_{HCO_3^{-}}=0.34mol/L*0.250L-0.0585mol=0.0265mol$

The resulting concentrations are:

$[CO_3^{2-}]=\frac{0.1435mol}{0.25L}=0.574M$

$[HCO_3^{-}]=\frac{0.0265mol}{0.25L}=0.106M$

Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:

$pH=10.2+log(\frac{0.574M}{0.106M} )\\\\pH=10.9$

Which makes sense since basic OH⁻ ions were added.

Regards!