Answer
The addition of $1.5g$ $HI$ would not exceed the buffer capacity;
Work Step by Step
1. Calculate the molar mass $(HI)$:
1.01* 1 + 126.9* 1 ) = 127.91g/mol
2. Calculate the number of moles $(HI)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{1.5}{ 127.91}$
$n(moles) = 0.0117$
3. Find the concentration in mol/L $(HI)$:
$0.0117$ mol in 1L: $0.0117 M (HI)$
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 = $[H_3O^+]$
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 acid, this reaction will occur:
$N{O_2}^-(aq) + H_3O^+(aq) -- \gt HNO_2(aq) + H_2O(l)$
And these are the concentrations after this reaction:
Remember: Reactants at equilibrium = Initial Concentration - y
And Products = Initial Concentration + y
Since $HI'$ is a strong base, y = $[HI'] = 0.0117M$
$[HNO_2] = 0.125 M + 0.0117 = 0.137M$
$[N{O_2}^-] = 0.145M - 0.0117 = 0.133M$
8. Now, calculate the hydronium ion concentration after the addition of the $HI'$:
$[H_3O^+] = Ka * (\frac{[HNO_2]}{[N{O_2}^-]})$
$[H_3O^+] = 4.6 \times 10^{-4} * \frac{0.137}{0.133}$
$[H_3O^+] = 4.6 \times 10^{-4} * 1.03$
$[H_3O^+] = 4.72 \times 10^{-4}$
9. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 4.72 \times 10^{- 4})$
$pH = 3.326$
10. Calculate the pH difference:
$3.401 - 3.326 = 0.075$
Since it is smaller than 1, we consider that, this addition did not exceed the buffer capacity;
