The OH⁻ concentration in a mixture of 5.0 mL of 0.0504 M NH₄Cl and 2 mL of 0.0210 M NH₃ solution is x × 10⁻⁶ M. The value of x is ___________. (Nearest integer)
[Given K w = 1 × 10⁻¹⁴ and K b = 1.8 × 10⁻⁵]

Solution

Chemistry ma’am · 4 steps

Count millimoles of the base and of its salt; the total volume cancels in their ratio.

1. Recognise the buffer. NH₃ is a weak base and NH₄Cl supplies its conjugate acid NH₄⁺. Together they form a basic buffer, so [OH⁻] comes straight from K_b and the base-to-salt ratio.

   [OH⁻] = K_b × n_(NH₃)/n_(NH₄⁺)

2. Find millimoles. NH₄Cl: 5.0 mL × 0.0504 M = 0.252 mmol. NH₃: 2 mL × 0.0210 M = 0.042 mmol. Both sit in the same 7 mL, so the volume cancels and you can use millimoles directly.
3. Put in numbers. The ratio is 0.042/0.252 = 1/6. So [OH⁻] = (1.8 × 10⁻⁵)/6 = 3 × 10⁻⁶ M.
4. So. [OH⁻] = 3 × 10⁻⁶ M, so x = 3.

The common slip: 8. Using the given molarities without multiplying by the volumes gives [OH⁻] = 1.8 × 10⁻⁵ × 0.0210/0.0504 = 7.5 × 10⁻⁶ M, which rounds to 8, not 3.

The idea to keep. In a buffer made by mixing two solutions, always work in moles (molarity times volume). The common final volume cancels, but the different starting volumes do not.

Answer from the NTA final answer key. Solution by PrepWiser.