Questions: In a solution, the concentration of KOH is 1.2 × 10^-3 M. Calculate the: a. [H^+]

In a solution, the concentration of KOH is 1.2 × 10^-3 M. Calculate the: a. [H^+]
Transcript text: In a solution, the concentration of KOH is $1.2 \times 10^{-3} \mathrm{M}$. Calculate the: a. $\left[\mathrm{H}^{+}\right]$
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Solution

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Solution Steps

Step 1: Determine the concentration of OH\(^-\)

Given the concentration of KOH is \(1.2 \times 10^{-3} \, \text{M}\), and since KOH is a strong base, it dissociates completely in water:

\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \]

Thus, the concentration of \(\text{OH}^-\) is also \(1.2 \times 10^{-3} \, \text{M}\).

Step 2: Use the water dissociation constant

The product of the concentrations of \(\text{H}^+\) and \(\text{OH}^-\) ions in water at 25°C is given by the water dissociation constant \(K_w\):

\[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \]

Step 3: Calculate the concentration of \(\text{H}^+\)

Rearrange the equation to solve for \([\text{H}^+]\):

\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} \]

Substitute the known values:

\[ [\text{H}^+] = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-3}} \]

Step 4: Perform the calculation

\[ [\text{H}^+] = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-3}} = 8.3333 \times 10^{-12} \]

Rounding to four significant digits:

\[ [\text{H}^+] = 8.333 \times 10^{-12} \, \text{M} \]

Final Answer

\(\boxed{[\text{H}^+] = 8.333 \times 10^{-12} \, \text{M}}\)

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