Titration Problems A sample of ammonium chloride (NH4Cl, MW = 53.49 g/mol; Ka, N

Question

Titration Problems

A sample of ammonium chloride (NH4Cl, MW = 53.49 g/mol; Ka, NH4 + = 5.70 x 10-10) was contaminated with inert material. The whole sample was dissolved in 25.00 mL of water, and this solution was titrated with 12.45 mL of 0.1000 M NaOH to reach the end point.

a. Calculate the amount of ammonium chloride present in the sample (in grams).

b. What was the pH of the solution at the beginning of the titration (assuming no contribution from the inert material).

c. What was the pH of the solution after the addition of 5.00 mL of 0.1000 M NaOH (assuming no contribution from the inert material)?

d. What was the pH of the solution at the equivalence point (assuming no contribution from the inert material)?

Thanks!

Explanation / Answer

a) nO Of mol of NaOH reacted = 12.45/1000*0.1   = 0.001245 mol

No of mol of NH4Cl presentin sample = 0.001245 mol

mass of nh4cl sample = 0.001245*53.491 = 0.066 grams

In the begining

concentration of NH4Cl = 0.001245/25*1000 = 0.0498 M

pka = -logka = -log(5.7*10^(-10)) = 9.24

b.   PKB OF nh3 = 4.74

pH = 7+1/2(4.74+log0.0498) = 8.72

c. pH = pka+log(salt/acid)

no of mol of NaOH = 5/1000*0.1 = 0.0005 mol

       = 9.24+log(0.0005/(0.001245-0.0005))
     = 9.066

d. at equivalence point

CONCENTRATION OF SALT = 0.001245/(25+12.5)*1000

poH = 1/2(pkB-logC)

     = 1/2(4.74-log0.0332)

pH = 10.9

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