$pH$ of mixtures

Tutorial 13

pH of a mixture of base and acid Simple Treatment If one of the two is strong (or both), - we write (no need for ionic writing!): The equation of their reaction which is complete in this case - we search: A) The numbers of moles before reaction B) The numbers of moles after reaction The pH is calculated according to the composition of the mixture after reaction (see examples) Complete treatment If both are weak ( see $ pKa_1 $ and $ pKa_2 $), a simple processing is not indicated!

What is the pH of a mixture of $ 1.0 \; L \; NaOH \ 0.15 \; \frac{mol}{L} $ with $ 1.0 \; L \; H_3PO_4 \; 0.10 \; \frac{mol}{L} $? Write the (nonionic) equation of the reaction!

For answers, use (possibly several times) the arrows ↑ Down! and ↓ Up! Complete please this question before moving on to the next one!

Complete the following table:

	$NaOH$	+	$H_3PO_4$		$NaH_2PO_4$	+	$H_2O$
initial nb of moles	0.15		0.10		0
variation of mole nbs	........		........		........
final nb of moles	........		........		........

	$NaOH$	+	$H_3PO_4$		$NaH_2PO_4$	+	$H_2O$
initial nb of moles	0,15		0,10		0
variation of mole nbs	-0.10		-0.10		+0.10
final nb of moles	0.05		0		0.10

$NaH_2PO_4$ dissociates to $H_2PO_4^-$and $Na^+$ $NaOH$ dissociates to $OH^-$ and $Na^+$ $Na^+$ has no influence. $H_2PO_4^-$ Is an ampholyte, but also a weak acid! $OH^-$ is treated as a strong base! There will be a reaction. Which?

$NaOH$ $+$ $NaH_2PO_4$ $ \rightarrow $ $Na_2HPO_4$ $+$ $H_2O$

Complete the following table:

	$NaOH$	+	$NaH_2PO_4$		$Na_2HPO_4$	+	$H_2O$
initial nb. of moles	0.05		0.10		0
variation of mole nbs	........		........		........
final nb. of moles	........		........		........

	$NaOH$	+	$NaH_2PO_4$		$Na_2HPO_4$	+	$H_2O$
initial nb. of moles	0,05		0,10		0
variation of mole nbs	-0.05		-0.05		+0.05
final nb. of moles	0.0		0.05		0.05

Write the solutes present at the end with their numbers of moles!

$n_{NaH_2PO_4}= 0.05\; mol$ $n_{Na_2HPO_4}= 0.05\; mol$

Analyse the final mixture!

$NaH_2PO_4$ dissociates to $H_2PO_4^-$ and $Na^+$ $Na_2HPO_4$ dissociates to $HPO_4^{2-}$ and $2Na^+$ $Na^+$ has no influence. $H_2PO_4^-$ and $HPO_4^{2-}$ are a buffer !

Calculate its pH!

$pH$ $=$ $pK_a+log\frac{n_{HPO_4^{2-}}}{n_{H_2PO_4^-}}$ $=$ $7.21+log\frac{0.05}{0.05}$ $=$ $7.21$ .