pH of HCI is 2.1. what is `[OH^(-)]` in its solution?

Options

A. `1.259 xx 10^(-12)` mol/I
B. `1.259 xx 10^(-12) g//dm^(3)`
C. `12.6 xx 10^(-12)` mol/I
D. `12.59 xx 10^(-12) g//dm^(3)`

Answer

To find the concentration of hydroxide ions \([OH^-]\) in a solution of hydrochloric acid (HCl) with a pH of 2.1, 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 the pH is 2.1, we can rearrange this equation to find \([H^+]\):

\[ [H^+] = 10^{-pH} = 10^{-2.1} \]

### Step 2: Calculate \(10^{-2.1}\)

Using a calculator or logarithmic tables, we can compute:

\[ 10^{-2.1} \approx 0.00794 \, \text{mol/L} \]

So,

\[ [H^+] \approx 7.94 \times 10^{-3} \, \text{mol/L} \]

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

At 25°C, the ion product of water (\(K_w\)) is given by:

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

Now, we can rearrange this equation to solve for \([OH^-]\):

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

Substituting the values we have:

\[ [OH^-] = \frac{1.0 \times 10^{-14}}{7.94 \times 10^{-3}} \]

### Step 4: Calculate \([OH^-]\)

Now, performing the division:

\[ [OH^-] \approx \frac{1.0 \times 10^{-14}}{7.94 \times 10^{-3}} \approx 1.26 \times 10^{-12} \, \text{mol/L} \]

### Final Answer

Thus, the concentration of hydroxide ions \([OH^-]\) in the solution is approximately:

\[ [OH^-] \approx 1.26 \times 10^{-12} \, \text{mol/L} \]

---