# Question 58460

Nov 23, 2015

${\text{245 g KNO}}_{2}$

#### Explanation:

First thing first, the $p {K}_{a}$ of the acid is constant as long as the acid dissociation constant, ${K}_{a}$, is constant, which is another way of saying that as long as temperature stays constant, the value of the acid's $p {K}_{a}$ will be constant as well.

More specifically, it will be equal to

$\textcolor{b l u e}{p {K}_{a} = - \log \left({K}_{a}\right)}$

$p {K}_{a} = - \log \left(4.5 \cdot {10}^{- 4}\right) = 3.35$

Now, the pH of a buffer solution that consists of a weak acid and its conjugate base can be calculated using the Henderson - Hasselbalch equaation, which looks like this

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))

An important thing to take away from this equation is that when you have equal concentrations of weak acid and conjugate base, the pH of the buffer will be equal to the acid's $p {K}_{a}$.

In your case, the pH of the buffer is higher than the $p {K}_{a}$ of the acid. This means that the concentration of the conjugate base must be higher than that of the weak acid.

What you have to do now is solve the H - H equation for the ratio that exists between the concentration of the conjugate base and that of the weak acid.

"pH" = pK_a + log( (["NO"_2^(-)])/(["HNO"_2]))

$3.88 = 3.35 + \log \left(\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right)\right)$

This is equivalent to

$\log \left(\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right)\right) = 0.53$

$\left(\left[{\text{NO"_2^(-)])/(["HNO}}_{2}\right]\right) = {10}^{0.53} = 3.39$

The concentration of the conjugate base will thus be

$\left[{\text{NO"_2^(-)] = 3.39 xx ["HNO}}_{2}\right]$

Since the molarity of the nitrous acid, ${\text{HNO}}_{2}$, is known to be $\text{1.70 M}$, the concentration of the nitrite anion, ${\text{NO}}_{2}^{-}$, the conjugate base of nitrous acid, will be

["NO"_2^(-)] = 3.39 xx "1.70 M" = "5.76 M"

Now, the nitrite anions are provided to the solution by dissolving potassium nitrite, ${\text{KNO}}_{2}$.

Since you were told that the volume of the solution does not change upon the addition of the salt, you can find the number of moles of nitrite anions needed by

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

$n = {\text{5.76 M" * "0.500 L" = "2.88 moles NO}}_{2}^{-}$

Potassium nitrite dissociates completely in aqueous solution to give potassium cations, ${\text{K}}^{+}$, and nitrite anions, ${\text{NO}}_{2}^{-}$

${\text{KNO"_text(2(aq]) -> "K"_text((aq])^(+) + "NO}}_{\textrm{2 \left(a q\right]}}^{-}$

You have a $1 : 1$ mole ratio between potassium nitrite and the nitrite anions, the number of moles of potassium nitrite will be

${n}_{K N {O}_{2}} = {n}_{N {O}_{2}^{-}} = {\text{2.88 moles KNO}}_{2}$

Finally, use potassium nitrite's molar mass to help you determine how many grams will contain this many moles

2.88color(red)(cancel(color(black)("moles KNO"_2))) * "85.104 g"/(1color(red)(cancel(color(black)("mole KNO"_2)))) = color(green)("245 g KNO"_2)#