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

The new pH is 10.8031

Step-by-step 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(moles)/(L) *1L=2moles`

Moles of HA:

`n_(HA) =2(moles)/(L) *1L=2moles`

Moles of NaOH:

`n_(NaOH) =15g*(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(molesNaA)/(molesHA) =10.6383+log(2.375)/(1.625) =10.8031`