A 5.43 g sample of NH CI was added to 24.5 mL of 1.11 M NaOH and the resulting solution diluted to 0.100 L. (Assume Kw...

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Question

A 5.43 g sample of NH CI was added to 24.5 mL of 1.11 M NaOH and the resulting solution diluted to 0.100 L. (Assume Kw = 1.01

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Answer 1

Answer #1

a) The moles of NH4 + and NaOH are calculated:

n NH4 = g / MM = 5.43 / 53.5 = 0.102 mol

n NaOH = M * V = 1.11 * 0.0245 L = 0.027 mol

NaOH reacts with NH4 + and forms NH3, the pH is calculated:

pH = pKa + log (n NH3 / n NH4 +) = 9.25 + log (0.027 / 0.102 - 0.027) = 8.81

b) It is a basic solution, the pH is greater than 7.

c) The moles of HCl added are calculated:

n HCl = 0.041 M * 0.00278 L = 0.00011 mol

The HCl reacts with the NH3 and forms NH4 +:

pH = 9.25 + log (0.027 - 0.00011 / 0.075 + 0.00011) = 8.80

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Answer 2

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