Question

What mass of KX (Mm= 170 g/mol) is required to prepare 385 mL of a pH = 12.87 solution? The pKa of HX is 12.50

g KX

Answer 1

Answer 2

1. Work Backwards from pH

• The solution is pH 12.87 (very basic).

• First, find pOH:

  $$\text{pOH}=14-\text{pH}=14-12.87=1.13$$

• Then, calculate the [OH⁻] concentration:

  $$[\text{OH⁻}]=10^{-\text{pOH}}=10^{-1.13}\approx 0.0741\text{M}$$

2. Connect [OH⁻] to the Salt (KX)

• KX dissolves into K⁺ + X⁻, and X⁻ reacts with water to make OH⁻:

  $$\text{X⁻}+\text{H₂O}\rightleftharpoons \text{HX}+\text{OH⁻}$$

• We need to find how much X⁻ is needed to produce 0.0741 M OH⁻.

• 
  

3. Find Kb (Base Strength of X⁻)

• Given pKa of HX = 12.50, we find Kb:

  $$\text{pKb}=14-\text{pKa}=14-12.50=1.50$$$$\text{Kb}=10^{-\text{pKb}}=10^{-1.50}\approx 0.0316$$

4. Calculate [X⁻] Needed

• Using the Kb expression:

  $$\text{Kb}=\frac{[\text{OH⁻}{]}^2}{[\text{X⁻}]}$$$$0.0316\approx \frac{(0.0741{)}^2}{[\text{X⁻}]}$$$$[\text{X⁻}]\approx \frac{0.00549}{0.0316}\approx 0.173\text{M}$$

5. Find Mass of KX

• For 385 mL (0.385 L):

  $$\text{Moles of KX}=0.173\text{M}\times 0.385\text{L}\approx 0.0146\text{moles}$$

• Convert to grams (since Mm = 170 g/mol):

  $$\text{Mass}=0.0146\text{moles}\times 170\text{g/mol}\approx 2.48\text{g}$$

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Final Answer:

Weigh out 2.48 grams of KX, dissolve it in water, and dilute to 385 mL to get a pH 12.87 solution.