pH values of solution x

Question

Please help me with this quesions! I feel I have an overall understanding of everything but I am having trouble applying it to these...

A student prepares a solution (Solution X) by dissolving 0.045 mol of HA and 0.072 mol of NaA into a beaker containing pure water.

1. NaA is fully soluble in water and forms Na⁺ (aq) and A^– (aq) ions.
2. If the total volume of solution X is 150.0 mL and the K_a of HA = 1.9 x 10^-6, calculate the pH value of Solution X at 25^oC.

Suppose we now add 0.020 mol of the strong acid HCℓ to Solution X.

1. How many moles of HA will be present at equilibrium after the HCℓ was added?
2. How many moles of A^– will be present at equilibrium after the HCℓ was added?

Answer 1

Since we are given a Ka for HA, we know it must be a weak acid. Thus, we have a weak acid plus the salt of that acid (NaA), which makes a BUFFER. When we have a buffer, we can use the Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

pKa = -log Ka = -log 1.9x10^-6 = 5.72

(1) pH = pKa + log [NaA] / [HA]

[NA] = 0.072 mol / 0.150 L = 0.480 M

[HA] = 0.045 mol / 0.150 L = 0.300 M

pH = 5.72 + log (0.480 / 0.300) = 5.72 + 0.20

pH = 5.92

(2) Adding 0.20 mol of HCl:

HCl will react with the A- thus reducing the amount of A- present and increasing the amount of HA present.

moles of H⁺ added = 0.020 moles

moles A- initially present = 0.072

moles HA initially present = 0.045

HA + A^- + HCl ==> HA + Cl^-

Final moles A^- = 0.072 mol - 0.020 mol = 0.052 mol

Final moles HA = 0.045 mol + 0.020 mol = 0.065 mol

If you prefer to use an ICE table, it might look something like this...

HA + A- + H+ ===> HA

0.045..0.072...0.02.........Initial

+0.02..-0.02...-0.02........Change

0.065...0..053....0.............Equilibrium