If 10.0 mL of 0.10 M HLac it's combined with 15.0 mL of 0.10 M NaLac...

If 10.0 mL of 0.10 M HLac it's combined with 15.0 mL of 0.10 M NaLac and 6.0 mL of 0.20 M NaOH, what is the final pH?

I have a hard time setting it up, can someone explain it from beginning to end?

Solutions

Expert Solution

The system contains 10.0 mL of 0.10 M HLac, 15.0 mL of 0.10 M NaLac and 6.0 mL of 0.20 M NaOH.

Millimols HLac = (10.0 mL)*(0.10 M) = 1.0 mmol

Millimols NaLac = (15.0 mL)*(0.10 M) = 1.5 mmol.

Millimols NaOH = (6.0 mL)*(0.20 M) = 1.2 mmol

Now, HLac is a weak acid and will react with a strong base like NaOH as per the reaction below.

HLac (aq) + NaOH (aq) ----------> NaLac (aq) + H₂O (l)

HLac is completely neutralized by NaOH to NaLac.

Millimols NaOH used to neutralize HLac = 1.0 mmol.

Millimols of excess NaOH in the system = (1.2 - 1.0) mmol = 0.2 mmol

The system therefore, contains a weak base, NaLac and a strong base, NaOH. Since excess NaOH is present in the system, hence, the pH of the system will be governed by the excess NaOH.

Total volume of the solution = (10.0 + 15.0 + 6.0) mL = 31.0 mL.

Concentration of excess NaOH in the solution = (0.2 mmol)/(31.0 mL) = 0.00645 M.

The pOH of the solution is given as

pOH = -log [OH^-]

= -log (0.00645 M)

= 2.19

It is known that

pH + pOH = 14.00

Therefore,

pH = 14.00 - pOH

= 14.00 - 2.19

= 11.81

≈ 11.8 (ans, correct to 3 sig. figs).

queen_honey_blossom answered 10 months ago

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