# Question #c8f95

##### 2 Answers
Feb 26, 2015

Since you're dealing with a buffer, you can use the Henderson-Hasselbalch equation to calculate the new pH of the solution when the concentrations of dihydrogen phosphate, ${H}_{2} P {O}_{4}^{-}$, and hydrogen phosphate, $H P {O}_{4}^{2 -}$, are equal.

The balanced chemical equations for this buffer are

${H}_{2} P {O}_{4}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + {H}_{3}^{+} O$, $p K {a}_{2} = 7.21$
$H P {O}_{4}^{2 -} + {H}_{2} O r i g h t \le f t h a r p \infty n s P {O}_{4}^{3 -} + {H}_{3}^{+} O$, $p K {a}_{3} = 12.7$

Equal concentrations of ${H}_{2} P {O}_{4}^{-}$ and $H P {O}_{4}^{2 -}$ will establish the first equilibrium, which implies that the pH of the solution will now be

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

$p {H}_{\text{solution") = 7.21 + log ("5.25 mmol/L"/"5.25 mmol/L}} = 7.21 + \log \left(1\right) = 7.21$

You can determine the GIbbs free energy by using the reaction's $p K {a}_{2}$ by using

$\Delta G = - R T \ln \left({K}_{a 2}\right)$, where

${K}_{a 2}$ is the acid dissociation constant for the established equilibrium reaction.

You can use the mathematical identity $\ln \left(x\right) = 2.303 \cdot \log \left(x\right)$ to rewrite the above equation as

$\Delta G = - R T \cdot 2.303 \log \left({K}_{a 2}\right)$

If you plug $p {K}_{a 2} = - \log \left({K}_{a 2}\right)$ into the equation, you'll get

$\Delta G = 2.303 R T \cdot p {K}_{a 2}$

Therefore,

$\Delta G = 2.303 \cdot 8.3145 \text{J"/("mol" * "K") * (273.15 + 25)"K} \cdot 7.21$

$\Delta G = \text{41,162.3 J/mol" = "+41.2 kJ/mol}$ $\to$ rounded to three sig figs.

SIDE NOTE. $\Delta G$ will be positive because the acid dissociation constant for the established equilibrium is smaller than 1.

Feb 27, 2015

$\Delta {G}_{r} = 41.12 k J . m o {l}^{- 1}$

$\Delta {G}_{r} = \Delta {G}^{0} + R T \ln Q$

We are concerned with:

${H}_{2} P {O}_{4}^{-} r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + {H}^{+}$

For which $p K {a}_{2} = 7.1$

$Q$ is the reaction quotient which, in this case $= \frac{5.25 m M}{5.25 m M} = 1$

$\Delta {G}_{r} = \Delta {G}^{0} + R T \ln Q$

Since $Q = 1$, $R T \ln Q = 0$

So $\Delta {G}_{r} = \Delta {G}^{0} = - R T \ln K {a}_{2}$

$\ln K {a}_{2} = 2.303 \log K {a}_{2}$

So $\Delta {G}_{r} = - 8.31 \times 298 \times 2.303 \times \left(- 7.21\right) = 41.12 k J . m o {l}^{- 1}$