A solution of acetic acid in water is adjusted to pH 5.01 with NaOH. What is the

Question

A solution of acetic acid in water is adjusted to pH 5.01 with NaOH. What is the ratio of prorogated acetic acid to detonated acetate? The pK_a of acetic acid is 4.75 If the solution described in #1 was initially prepared by making a 5% solution of solid acetic acid (w/v) in water before pH adjustment, what is the actual concentration of acetic acid (the prorogated form only) after pH adjustment? What is the pH at which the ratio of lactate:lactic acid in solution is 1.5:7? The pKa of lactic acid is 3.86.

Explanation / Answer

1.

pH = pka + log(salt/acid)

5.01 = 4.75+log(x)

x = salt/acid = 1.82

2. 5% solution. means 5 grams in 100 ml solution.

No of mol of aceticacid initial = 5/60 = 0.083 mol

1.82 = x/(0.083-x)

x = 0.0535 mol

so that,

No of mol of aceticacid after = 0.083-0.0535

= 0.0295 mol

mass of aceticacid = (0.0295*60) = 1.77 grams

concentration of acid = 1.77%

3. pH = 3.86+log(1.5/7) = 3.19

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