What is `[H^(+)]` in a solution that is 0.01 M in HCn and 0.02 M in NaCN ?
`(K_(a)` for HCN ` 6.2 xx 10^(-10))`

Options

A. `3.1 xx 10^(10)`
B. `6.2 xx 10^(5)`
C. `6.2 xx 10^(-10)`
D. `3.1 xx 10^(-10)`

Answer

To find the concentration of \([H^+]\) in a solution that is 0.01 M in HCN and 0.02 M in NaCN, we can follow these steps:

### Step 1: Understand the dissociation of NaCN
NaCN dissociates completely in solution:
\[ \text{NaCN} \rightarrow \text{Na}^+ + \text{CN}^- \]
From this reaction, we can see that the concentration of \(\text{CN}^-\) ions will be equal to the concentration of NaCN, which is 0.02 M.

### Step 2: Write the dissociation reaction for HCN
The dissociation of HCN in water can be represented as:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]

### Step 3: Set up the expression for \(K_a\)
The acid dissociation constant (\(K_a\)) for HCN is given by:
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]
Given that \(K_a = 6.2 \times 10^{-10}\), we can substitute the known concentrations into this expression.

### Step 4: Substitute known values into the \(K_a\) expression
Let \([H^+]\) be \(x\). The concentration of \(\text{CN}^-\) is already known to be 0.02 M, and the initial concentration of HCN is 0.01 M. Since HCN is a weak acid, we can assume that the change in concentration of HCN due to dissociation is negligible compared to its initial concentration. Thus:
\[ K_a = \frac{x \cdot 0.02}{0.01} \]

### Step 5: Solve for \([H^+]\)
Substituting the values into the equation:
\[ 6.2 \times 10^{-10} = \frac{x \cdot 0.02}{0.01} \]
This simplifies to:
\[ 6.2 \times 10^{-10} = 2x \]
Now, solving for \(x\):
\[ x = \frac{6.2 \times 10^{-10}}{2} = 3.1 \times 10^{-10} \]

### Conclusion
Thus, the concentration of \([H^+]\) in the solution is:
\[ [H^+] = 3.1 \times 10^{-10} \, \text{M} \]