2) A 350.0 mL buffer solution is 0.140 M in HFand 0.140 M in NaF. What mass of N

Question

2) A 350.0 mL buffer solution is 0.140 M in HFand 0.140 M in NaF. What mass of NaOH could this buffer neutralize before the pH rises above 4.00? If the same volume of the buffer was 0.360 M in HF and 0.360 M in NaF, what mass of NaOHcould be handled before the pH rises above 4.00?

3) A 5.65 g sample of a weak acid with Ka=1.3×104 was combined with 5.20 mL of 6.10 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.25.What is the molar mass of the weak acid?

Explanation / Answer

2) pH of acidic buffer = pka + log(NaF/HF)

pka of HF = 3.17

       = 3.17+log(0.14/0.14)

pH = 3.17

after addition of NaOH

4 = 3.17+LOG((0.36+X)/(0.36-X))

X = 0.267 mol

No of mol of NaOH = 0.267 mol

mass of NaOH = 0.267*40 = 10.68 grams

3)

pka of weak acid = -log(1.3*10^(-4)) = 3.88

No of mol of NaOH added = 5.2/1000*6.1 = 0.0317 mol

No of mol of ACID = ?

4.25 = 3.88 + LOG(0.0317/(X-0.0317))

X = No of mol of ACID = 0.0452 mol

Molarmass of acid = 5.65/0.0452 = 125 g/mol

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