# Is this enough information to solve this problem?

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 $\mathrm{pH=3.35}$.

Since this is a titration problem shouldn't I be given one of the volumes ?

• If you're given the volume of one of the species as $x$ (say) then all you've to do to find the other species' volume is $300-x$. Then question, then, would be meaningless. – Berry Holmes May 26 '17 at 3:32

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,

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

• You can use $\ce{chemical formula goes here}$ for writing compounds $\pu{units here} for physical units and \Rightarrow for$\Rightarrow\$. – Berry Holmes May 26 '17 at 4:46