A solution of an acid has `pH= 4.70`. Find out the concentration of `OH^(-)` ions `(pK_(w)=14)`.

Options

A. `5xx10^(-10)M`
B. `4xx10^(-10)M`
C. `2xx10^(-5)M`
D. `9xx10^(-10)M`

Explanation

To find the concentration of hydroxide ions \([OH^-]\) in a solution with a given pH of 4.70, we can follow these steps:

### Step 1: Calculate the concentration of hydrogen ions \([H^+]\)

The pH of a solution is related to the concentration of hydrogen ions by the formula:
\[ pH = -\log[H^+] \]
Given that \(pH = 4.70\), we can rearrange the formula to find \([H^+]\):
\[ [H^+] = 10^{-pH} = 10^{-4.70} \]

### Step 2: Calculate \([H^+]\)

Using a calculator or logarithmic tables:
\[ [H^+] = 10^{-4.70} \approx 1.995 \times 10^{-5} \text{ M} \]
For simplicity, we can round this to:
\[ [H^+] \approx 2.00 \times 10^{-5} \text{ M} \]

### Step 3: Use the ion product of water to find \([OH^-]\)

The ion product of water (\(K_w\)) at 25°C is:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
We can rearrange this equation to solve for \([OH^-]\):
\[ [OH^-] = \frac{K_w}{[H^+]} \]

### Step 4: Substitute the values into the equation

Now substituting the values we have:
\[ [OH^-] = \frac{10^{-14}}{2.00 \times 10^{-5}} \]

### Step 5: Calculate \([OH^-]\)

Performing the division:
\[ [OH^-] = 0.5 \times 10^{-9} \text{ M} = 5.0 \times 10^{-10} \text{ M} \]

### Final Answer

The concentration of hydroxide ions \([OH^-]\) in the solution is:
\[ [OH^-] \approx 5.0 \times 10^{-10} \text{ M} \]

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