Question

A 350.0 mLmL buffer solution is 0.150 molL−1molL−1 in HFHF and 0.150 molL−1molL−1 in NaFNaF.
Ka(HF)=6.3×10−4

If the same volume of the buffer was 0.370 molL−1molL−1 in HFHF and 0.370 molL−1molL−1 in NaFNaF, what mass of NaOHNaOH could be handled before the pHpH rises above 4.00?

Express your answer using two significant figures.

Answer 1

concentration of HF = 0.150 M

concentration of NaF = 0.150 M

volume of the buffer =350 mL

Ka of HF = 6.3 x 10^-4

pK_a = - log K_a

= - log (6.3 x 10^-4)

= 3.2

pH = pK_a + log ([base] / [acid])

in the above case as base and acid are having same concentration pH = pK_a

initial pH = 3.2

the number of moles of HF = 0.350 L x 0.150 mol/L

= 0.0525 mol

number of moles of NaF = 0.350 L x 0.150 mol/L

= 0.0525 mol

on adding NaOH

OH^- + HF --------> F^- + H2O

before reaction 0.0525 moles

addition x moles

after addition (0.0525 -x ) moles (0.0525 + x) moles

4 = 3.2 + log ((0.0525 + x) / (0.0525 -x ))

6.31 = ((0.0525 + x) / (0.0525 -x ))

x=0.0381

number of moles of NaOH is 0.0381

molar mass of NaOH = 39.996 g/mol

mass of NaOH = 0.0381 moles x 39.996 g/mol

= 1.525 g

concentration of HF = 0.370 M

concentration of NaF = 0.370 M

volume of the buffer =350 mL

Ka of HF = 6.3 x 10^-4

pK_a = - log K_a

= - log (6.3 x 10^-4)

= 3.2

pH = pK_a + log ([base] / [acid])

in the above case as base and acid are having same concentration pH = pK_a

initial pH = 3.2

the number of moles of HF = 0.350 L x 0.370 mol/L

= 0.1295 mol

number of moles of NaF = 0.350 L x 0.370 mol/L

= 0.1295 mol

on adding NaOH

OH^- + HF --------> F^- + H2O

before reaction 0.1295 moles

addition x moles

after addition (0.1295 -x ) moles (0.1295 + x) moles

4 = 3.2 + log ((0.1295 + x) / (0.1295 -x ))

6.31 = ((0.1295 + x) / (0.1295 -x ))

x=0.0941

number of moles of NaOH is 0.0941

molar mass of NaOH = 39.996 g/mol

mass of NaOH = 0.0941 moles x 39.996 g/mol

= 3.76 g