Answer
The $NaOH$ didn't exceed the capacity of the buffer.
Work Step by Step
1. Calculate the molar mass ($NaOH$):
22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol
2. Calculate the number of moles ($NaOH$)
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{0.25}{ 40}$
$n(moles) = 6.3\times 10^{- 3}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 6.3\times 10^{- 3}}{ 0.5} $
$C(mol/L) = 0.013 ($NaOH$)$
4. Drawing the ICE table we get these concentrations at the equilibrium:
$HNO_2(aq) + H_2O(l) \lt -- \gt N{O_2}^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[HNO_2] = 0.1 M - x$
$[N{O_2}^-] = 0.15M + x$
$[H_3O^+] = 0 + x$
5. Calculate 'x' using the $K_a$ expression.
$ 4.6\times 10^{- 4} = \frac{[N{O_2}^-][H_3O^+]}{[HNO_2]}$
$ 4.6\times 10^{- 4} = \frac{( 0.15 + x )* x}{ 0.1 - x}$
Considering 'x' has a very small value.
$ 4.6\times 10^{- 4} = \frac{ 0.15 * x}{ 0.1}$
$ 4.6\times 10^{- 4} = 1.5x$
$\frac{ 4.6\times 10^{- 4}}{ 1.5} = x$
$x = 3.1\times 10^{- 4}$
Percent dissociation: $\frac{ 3.1\times 10^{- 4}}{ 0.1} \times 100\% = 0.31\%$
x = $[H_3O^+]$
$[HNO_2] = 0.1 M - x = 0.1 M - 3.1 \times 10^{-4}M \approx 0.1M$
$[N{O_2}^-] = 0.15M + x = 0.1 M + 3.1 \times 10^{-4}M \approx 0.15M$
$[H_3O^+] = 0 + x = 3.1 \times 10^{-4}M$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 3.1 \times 10^{- 4})$
$pH = 3.51$
7. Since we are adding a strong base, this reaction will occur:
$HNO_2(aq) + OH^-(aq) -- \gt N{O_2}^-(aq) + H_2O(l)$
And these are the concentrations after this reaction:
Remember: Reactants at equilibrium = Initial Concentration - y
And Products = Initial Concentration + y
Since $NaOH$ is a strong base, y = $[NaOH] = 0.0125M$
$[HNO_2] = 0.1 M - 0.0125 = 0.088M$
$[N{O_2}^-] = 0.15M + 0.0125 = 0.16M$
8. Now, calculate the hydronium ion concentration after the addition of the $NaOH$:
$[H_3O^+] = Ka * (\frac{[HNO_2]}{[N{O_2}^-]})$
$[H_3O^+] = 4.6 \times 10^{-4} * \frac{0.088}{0.16}$
$[H_3O^+] = 4.6 \times 10^{-4} * 0.54$
$[H_3O^+] = 2.5 \times 10^{-4}$
9. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.5 \times 10^{- 4})$
$pH = 3.60$
10. Calculate the pH difference:
$3.60 - 3.51 = 0.09$
Since it is smaller than 1, we consider that, this compound did not exceed the buffer capacity.
