Question

One liter of a buffer contains 2.00 Molar NaA and 2.00 Molar HA. (A- is the anion of an acid) 15.00 g NaOH is added. (Assume no volume change) What is the new pH?
HA (aq) + H2O (l) A - (aq) + H3O+ (aq) Ka = 2.3 x 10-11

Answer 1

Answer:

The new pH is 10.8031

Explanation:

Given information:

[NaA] = concentration = 2 M

[HA] = concentration = 2 M

V = volume = 1 L

15 g of NaOH added

Question: What is the new pH, pH = ?

Moles of NaA:

$n_{NaA} =2\frac{moles}{L} *1L=2moles$

Moles of HA:

$n_{HA} =2\frac{moles}{L} *1L=2moles$

Moles of NaOH:

$n_{NaOH} =15g*\frac{1mol}{40g} =0.375moles$

The reaction:

HA + NaOH → NaA + H₂O

Moles of NaA = 2 + 0.375 = 2.375 moles

Moles of HA = 2 - 0.375 = 1.625 moles

The value of Ka is 2.3x10⁻¹¹

The pKa:

$pKa=-logKa=-log(2.3x10^{-11} )=10.6383$

Applying the Henderson's Hasselbach equation

$pH=pKa+log\frac{molesNaA}{molesHA} =10.6383+log\frac{2.375}{1.625} =10.8031$