Answer
The addition of $NaOH$ $1.5g$ would not exceed the buffer capacity;
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 $(NaOH)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{1.5}{ 40}$
$n(moles) = 0.0375$
3. Find the concentration in mol/L $(NaOH)$:
$0.0375$ mol in 1L: $0.0375 M (NaOH)$
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
$[HNO_2] = 0.125 M - x$
$[N{O_2}^-] = 0.145M + x$
$[H_3O^+] = 0 + x$
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{( 0.145 + x )* x}{ 0.125 - x}$
Considering 'x' has a very small value.
$ 4.6\times 10^{- 4} = \frac{ 0.145 * x}{ 0.125}$
$ 4.6\times 10^{- 4} = 1.16x$
$\frac{ 4.6\times 10^{- 4}}{ 1.16} = x$
$x = 3.97\times 10^{- 4}$
Percent dissociation: $\frac{ 3.97\times 10^{- 4}}{ 0.125} \times 100\% = 0.317\%$
x = $[H_3O^+]$
$[HNO_2] = 0.125 M - x = 0.125 M - 3.97 \times 10^{-4}M \approx 0.125M$
$[N{O_2}^-] = 0.145M + x = 0.125 M + 3.97 \times 10^{-4}M \approx 0.145M$
$[H_3O^+] = 0 + x = 3.97 \times 10^{-4}M$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 3.97 \times 10^{- 4})$
$pH = 3.401$
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 = $[NaOH] = 0.0375M$
$[HNO_2] = 0.125 M - 0.0375 = 0.0875M$
$[N{O_2}^-] = 0.145M + 0.0375 = 0.183M$
8. Now, calculate the hydronium ion concentration after the addition of the $NaOH$:
$[H_3O^+] = Ka * (\frac{[HNO_2]}{[N{O_2}^-]})$
$[H_3O^+] = 4.7 \times 10^{-4} * \frac{0.0875}{0.183}$
$[H_3O^+] = 4.7 \times 10^{-4} * 0.479$
$[H_3O^+] = 2.21 \times 10^{-4}$
9. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.21 \times 10^{- 4})$
$pH = 3.656$
10. Calculate the pH difference:
$3.656 - 3.401 = 0.255$
Since it is smaller than 1, we consider that, the base did not exceed the buffer.
