# What is the pH of a buffer composed of 0.145*mol*L^-1 NaNO_2 and 0.125*mol*L^-1 HNO_2? How does the pH evolve if 2.75*g of HCl, and we ignore any volume change?

Jul 16, 2017

See here for the background.....

#### Explanation:

A buffer solution consists of appreciable concentrations of a weak acid and its conjugate base. Because the acid and conjugate base compete for the proton, the $p H$ of the buffer remains tolerably close to the $p {K}_{a}$ of the acid........

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

You can read up on how this formula is derived, but the missing datum from your question is $p {K}_{a}$ for $H N {O}_{2}$; this link tells me that $p {K}_{a} = 3.398$, and we put these data into the equation:

$p H = 3.398 + {\log}_{10} \left[\frac{0.145 \cdot m o l \cdot {L}^{-} 1}{0.125 \cdot m o l \cdot {L}^{-} 1}\right]$

Here of course, $\text{nitrous acid} = H N {O}_{2} \equiv H A$

And we plugs in the number and gets......$p H = 3.46$, which makes sense given that the base component is slightly in excess. Does this make sense to you........?

And now we add a $2.75 \cdot g$ mass of $H C l$, for which WE ASSUME NO VOLUME CHANGE. This represents a molar quantity of $\frac{2.75 \cdot g}{36.45 \cdot g \cdot m o {l}^{-} 1} = 0.0754 \cdot m o l$......

This protonates the nitrite ion, and thus slightly increases $\left[H N {O}_{2}\right]$ according to the following reaction......

$N a N {O}_{2} \left(a q\right) + H C l \left(a q\right) \rightarrow N a C l \left(a q\right) + H N {O}_{2} \left(a q\right)$....

And thus the concentration of $H N {O}_{2}$ slightly increases, and the $p H$ of the buffer SLIGHTLY decreases.........

$p H = 3.398 + {\log}_{10} \left\{\frac{0.145 \cdot m o l \cdot {L}^{-} 1}{0.125 + 0.0754 \cdot m o l \cdot {L}^{-} 1}\right\}$

$= 3.398 - 0.1405 = 3.257$.

Are you happy with this? Please check for errata........all care taken but no responsibility admitted.........