# 0.100M NaOH is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH at 4 different points in the titration? a.) Initial pH of acid b.) After 40.00 mL of NaOH is added c.) After 50.00 mL of NaOH is added d.) After 50.20 mL of NaOH has been added

Mar 16, 2015

In order, the pH at those specific four points in the titration will be

a) 1, b) 1.95, c) 7, d) 10.3

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

Notice that you have a $\text{1:1}$ mole ratio between sodium hydroxide and hydrochloric acid, i.e. it will take 1 mole of the former to neutralize 1 mole of the latter.

You start with the hydrochloric acid solution. It's pH will be determined by the concentration of hydronium ions present in solution. Because $H C l$ is a strong acid, it will dissociate completely in aqueous solution to give 1 mole of chloride ions and 1 mole of hydronium ions

$H C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} \to C {l}_{\left(a q\right)}^{-} + {H}_{3} {O}_{\left(a q\right)}^{+}$

Therefore,

pH_("initial") = -log([H_3O^(+)]) = -log(0.100) = "1"

Now you start adding the strong base. After you've added 40.0 mL of $N a O H$, the number of moles of hydrochloric acid left in solution will decrease because of the neutralization reaction that takes place between the strong acid and the strong base.

The number of moles of $H C l$ will decrease by the same amount of moles of $N a O H$ added.

$C = \frac{n}{V} \implies {n}_{\text{NaOH}} = C \cdot V$

n_("NaOH") = = "0.100 M" * 40.0 * 10^(-3)"L" = "0.004 moles NaOH"

The number of moles of hydrochloric acid in excess will be

n_("HCl excess") = n_("HCl") - n_("NaOH") = 0.005 - 0.004 = "0.001 moles"

The volume of the solution is now

V_("solution") = "50.0 mL" + "40.0 mL" = "90.0 mL"

The concentration of hydronium ions will be

C = n/V_("solution") = "0.001 moles"/(90.0 * 10^(-3)"L") = "0.0111 M"

As a result, the pH will now be

pH_("solution 1") = -log(0.0111) = "1.95"

Now you increase the volume of $N a O H$ solution added to 50.0 mL. The number of moles of sodium hydroxide added will be equal to the number of moles of hydrochloric acid present in the solution, which means that complete neutralization takes place.

At this point, kown as the equivalence point, no moles of the acid and of the base remain in solution. As a result, the pH of the solution will be neutral

pH_("equivalence point") = "7"

FInally, you add a little more base, 0.20 mL to be precise, to the now-neutral solution. This means that the concentration of the hydroxide ions will determine the pH of the solution.

The number of moles of $N a O H$ added is

$n = C \cdot V = \text{0.100 M" * 0.20 * 10^(-3)"L" = "0.00002 moles NaOH}$

The total volume of the solution will be

V_("total") = "50.0 mL" + "50.20 mL" = "100.2 mL"

The concentration of the hydroxide anions will be

C = n/V_("total") = "0.00002 moles"/(100.2 * 10^(-3)"L") = "0.0001996 M"

The pH of the final solution will be

$p {H}_{\text{final}} = 14 - p O H = 14 - \left(- \log \left(0.0001996\right)\right)$

pH_("final") = 14 - 3.7 = 10.3"