# Question 9c4f9

Jan 31, 2015

The pH of the phosphate buffer will be $\text{6.39}$.

The reactions that concern you are

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

So, adding monosodium phosphate will increase the concentration of ${H}_{2} P {O}_{4}^{-}$, while adding dosodium phosphate will increase the concentration of $H P {O}_{4}^{2 -}$.

The first thing you need to do is figure out the concentrations of the two species in the new solution. Starting with monosodium phosphate

n_("mono") = C * V = "80.0 mmol/L" * 250 * 10^(-3)"L" = "20.0 mmol"

The volume of the new solution will be

V_("sol") = "250 mL" + "250 mL" = "500 mL", which means that the concentration of monosodium phosphate (and of ${H}_{2} P {O}_{4}^{-}$) will be

C_("mono") = n_("mono")/V_("sol") = ("20.0 mmol")/(500*10^(-3)"L") = "40.0 mmol/L"

Now for the disodium phosphate

n_("di") = C * V = "30.0 mmol/L" * 250*10^(-3)"L" = "7.50 mmol"

The concentration of disodium phosphate (and of $H P {O}_{4}^{2 -}$) in solution will be

C_("di") = n_("di")/V_("sol") = ("7.50 mmol")/(500*10^(-3)"L") = "15.0 mmol/L"

Now use the Hendesron-Hasselbalch equation to determine the pH

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

pH_("sol") = 6.82 + log(("15.0 mmol/L")/("40.0 mmol/L"))#

$p {H}_{\text{sol}} = 6.82 - 0.426 = 6.39$