Question

determine the concentration of hydroxide ion, oh−, in a particular solution of bro− by constructing an ice table, writing the equilibrium constant expression, and using this information to determine the concentration of hydroxide ion. the kb for bro− is 4.0×10−6. complete parts 1 - 3 before submitting your answer. a 0.724 m aqueous solution of bro− was prepared. fill in the ice table with the appropriate value for each involved species to determine concentrations of all reactants and products.

Explanation:

Step1: Write the equilibrium reaction

The reaction for the hydrolysis of \( \text{BrO}^- \) is \( \text{BrO}^-(aq)+\text{H}_2\text{O}(l)
ightleftharpoons\text{OH}^-(aq)+\text{HBrO}(aq) \).

Step2: Set up the ICE table

• Initial (M):
• \( [\text{BrO}^-]=0.724\ \text{M} \), \( [\text{OH}^-]=0\ \text{M} \), \( [\text{HBrO}]=0\ \text{M} \) (since no products are formed initially).
• Change (M):
• Let \( x \) be the change in concentration. For \( \text{BrO}^- \), the change is \( -x \) (it is consumed). For \( \text{OH}^- \) and \( \text{HBrO} \), the change is \( +x \) (they are produced).
• Equilibrium (M):
• \( [\text{BrO}^-]=0.724 - x \), \( [\text{OH}^-]=x \), \( [\text{HBrO}]=x \)

Step3: Write the \( K_b \) expression

\( K_b=\frac{[\text{OH}^-][\text{HBrO}]}{[\text{BrO}^-]} \). Given \( K_b = 4.0\times10^{-6} \), substitute the equilibrium concentrations: \( 4.0\times10^{-6}=\frac{x\cdot x}{0.724 - x} \).
Since \( K_b \) is very small (\( 4.0\times10^{-6}\ll1 \)), \( 0.724 - x\approx0.724 \).
So, \( 4.0\times10^{-6}=\frac{x^{2}}{0.724} \).
Solve for \( x \):

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Answer:

The concentration of \( \text{OH}^- \) is \( 1.7\times 10^{-3}\ \text{M} \).