# Question 8fccb

Apr 13, 2015

So, you're dealing with a buffer that contains monosodium phosphate, $N a {H}_{2} P {O}_{4}$, and disodium phosphate, $N {a}_{2} H P {O}_{4}$.

In aquous solution, the species that are of interest for your buffer are dihydrogen phosphate, ${H}_{2} P {O}_{4}^{-}$, which will act as a weak acid, and hydrogen phosphate, $H P {O}_{4}^{2 -}$, which will act as the conjugate base.

Since nothing is actually added to the buffer, you have no net ionic equation to write. Now, because phosphoric acid is a triprotic acid, it will dissociate in three steps according to the following equilibrium reactions

${H}_{3} P {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{2} P {O}_{4 \left(a q\right)}^{-} + {H}_{3} {O}_{\left(a q\right)}^{+}$, $\text{ } p {K}_{a 1}$

${H}_{2} P {O}_{4 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H P {O}_{4 \left(a q\right)}^{2 -} + {H}_{3} {O}_{\left(a q\right)}^{+}$, $\text{ } p {K}_{a 2}$

$H P {O}_{4 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s P {O}_{4 \left(a q\right)}^{3 -} + {H}_{3} {O}_{\left(a q\right)}^{+}$, $\text{ } p {K}_{a 3}$

Notice that your buffer contains ${H}_{2} P {O}_{4}^{-}$ and $H P {O}_{4}^{2 -}$, which means that the second equilibrium reaction will be established.

As a result, ${K}_{a 2}$ will be the more important acid dissociation constant for your buffer.

Notice that the concentration of $H P {O}_{4}^{2 -}$ is 2 times bigger than the concentration of ${H}_{2} P {O}_{4}^{-}$; this means that the dominant form of the acid will be $H P {O}_{4}^{2 -}$ and that the solution's pH will be bigger than $p {K}_{a 2}$.

Moreover, because the difference between ${K}_{a 2}$ and ${K}_{a 3}$ is so significant, the concentration of $P {O}_{4}^{3 -}$ can be neglected.

So, since this is a buffer, you can use the Henderson-Hasselbalch equation to solve for the pH. Since the second equilibrium is set, you'll use $p {K}_{a 2}$, which is equal to

$p {K}_{a 2} = - \log \left({K}_{a 2}\right) = - \log \left(6.3 \cdot {10}^{- 8}\right) = 7.20$

Therefore,

$p {H}_{\text{sol}} = p {K}_{a 2} + \log \left(\frac{\left[H P {O}_{4}^{2 -}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]}\right)$

pH_"sol" = 7.20 + log((0.29cancel("M"))/(0.13cancel("M"))) = 7.20 + 0.35 = color(green)(7.55)#