We have (a) 50.0 mL of 0.10 M NH3 solution, (b) 50.0 mL of 0.12 M...

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Question

We have (a) 50.0 mL of 0.10 M NH3 solution, (b) 50.0 mL of 0.12 M HNO3 solution, (c) 50.0 mL of 0.12 M NaOH solution, and (d) 50.0 mL of 0.06 M HClO4 solution. Can you make a buffer by combining two of these solutions? If so, which ones? What would be the pH of the buffer solution? Please explain

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Answer 1

Answer #1

By combining the solutions of the weak base NH3 (a) and the strong acid HClO4 (d) a buffer is formed.

HClO4 reacts with NH3 (decreasing it) and NH4 + is formed (increasing it), the initial moles of NH3 and HClO4 are calculated:

n NH3 = M * V = 0.1 M * 0.05 L = 0.005 mol

n NH4 + = 0.06 M * 0.05 L = 0.003 mol

The final moles of NH3 are calculated:

n NH3 = 0.005 - 0.003 = 0.002 mol

The pH is calculated:

pH = pKa + log n Salt / n Acid = 9.24 + log (0.002 / 0.003) = 9.06

Answer 2

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