Question

# 2) A 350.0 −mL buffer solution is 0.140 M in HFand 0.140 M in NaF. What...

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=1.3×10−4 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?

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

pka of HF = 3.17

= 3.17+log(0.14/0.14)

pH = 3.17

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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