23 and b) I just want to see how they got that. A sample of ammonium chloride (N

Question

23 and b) I just want to see how they got that.

A sample of ammonium chloride (NH_4Cl, MW = 53.49 g/mol, Ka, NH_4^+ = 5.70 times 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 of0.1000 M NaOH to reach the end point. Calculate the amount of ammonium chloride present in the sample (in grams). What was the pH of the solution at the beginning of the titration (assuming no contribution from the inert material).

Explanation / Answer

a. moles of NaOH used = 0.1 M x 12.45 ml = 1.245 mmol

1 mole of NaOH reacts with 1 mole of NH4Cl

moles of NaOH = moles of NH4Cl

Amount of NH4Cl present in sample = 1.245 mmol x 53.491 g/mol/1000 = 0.0666 g

b. NH4+ <==> NH3 + H+

let x amount has dissociated

Ka = 5.70 x 10^-10 = x^2/0.05

x = [H+] = 5.34 x 10^-6 M

pH = -log[H+] = 5.27

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