Is this enough information to solve this problem?


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> Calculate the volume of a $\pu{0.100M}$ solution of $\ce{HNO2}$ and the volume of a $\pu{0.200M}$ solution of $\ce{NaNO2}$, that must be mixed to prepare $\pu{300ml}$ of a solution with a $\mathrm{pH=3.35}$.
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Since this is a titration problem shouldn't I be given one of the volumes ?




You need to use the Henderson-Hasselbach equation:


$$\pu{pH = pK\_a + \log\frac{[Base]}{[Acid]}}$$


The $\pu{pK\_a}$ of $\ce{HNO2}= 3.15$


Therefore,


$$10^{\pu{pH-pK\_a}} = \pu{\frac{[Base]}{[Acid]}}$$
$$\pu{10^{3.35-3.15} = \frac{[Base]}{[Acid]}}$$
$$\pu{1.58 = \frac{[Base]}{[Acid]}}$$


Implies that the numerator must be $1.58$ times that of the denominator.


The base is $\ce{NaNO2}$; the acid is $\ce{HNO2}$. The base is two times as concentrated as the acid.


This means, for every $\pu{mL}$ of acid you use, you need $\pu{1.58/ 2 = 0.79 mL}$ of base.


If,


$$V\_T = V\_A + V\_B$$


And,


$$V\_B = 0.79V\_A$$


Then,


$$\pu{300 mL} = V\_A + 0.79V\_A \Rightarrow V\_A = \pu{167.60 mL}$$


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