# Question b3ad7

May 14, 2017

Warning! Very long Answer. You have to calculate the pH at a number of points before and after the equivalence point and then plot pH vs volume.

#### Explanation:

The calculations are tedious but not difficult.

The balanced equation is

$\text{NaOH + HCl" → "NaCl" + "H"_2"O}$

pH at start

["OH"^"-"] = "0.007 25 mol/L"

"pOH" = "-log"("0.007 25") = 2.14

$\text{pH = 14.00 - pOH = 14.00 - 2.14 = 11.86}$

pH at equivalence point

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {c}_{\textrm{A}} {V}_{\textrm{A}} = {c}_{\textrm{B}} {V}_{\textrm{B}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

V_text(A) = V_text(B) × c_text(B)/c_text(A) = "690 mL" × ("0.007 25" color(red)(cancel(color(black)("mol/L"))))/(0.10 color(red)(cancel(color(black)("mol/L")))) = "50.0 mL"

The equivalence point is at 50.0 mL $\text{HCl}$.

pH at 25 mL

At this point, you have neutralized half the moles of $\text{NaOH}$.

$\text{Initial moles" = 0.690 color(red)(cancel(color(black)("L NaOH"))) × ("0.007 25 mol NaOH")/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.005 00 mol}$

$\text{Moles present" = 0.5 × "0.005 00 mol" = "0.002 500 mol}$

$\text{Volume = (690 + 25) mL = 715 mL = 0.715 L}$

["OH"^"-"] = "0.002 500 mol"/"0.715 L" = "0.003 50 mol/L"

$\text{pOH = -log(0.003 50) = 2.46}$

$\text{pH = 14.00 - 2.46 = 11.54}$

pH at 49 mL

$\text{Moles reacted" = 49/50 × "0.005 00 mol" = "0.004 90 mol}$

$\text{Moles present" = "(0.005 00 -0.004 90) mol" = "0.000 10 mol}$

$\text{Volume = (690 + 49) mL = 739 mL = 0.739 L}$

["OH"^"-"] = "0.000 10 mol"/"0.739 L" = "0.000 135 mol/L"

$\text{pOH = -log(0.000 135) = 3.87}$

$\text{pH = 14.00 - 3.87 = 10.13}$

pH at 49.9 mL

$\text{Moles reacted = 49.9/50.0 × "0.005 00 mol" = "0.004 99 mol}$

$\text{Moles present" = "(0.005 00 -0.004 99) mol" = "0.000 01 mol}$

$\text{Volume = (690 + 49.9) mL = 739.9 mL = 0.7399 L}$

["OH"^"-"] = "0.000 01 mol"/"0.7399 L" = "0.000 014 mol/L"#

$\text{pOH = -log(0.000 014) = 4.87}$

$\text{pH = 14.00 - 4.87 = 9.13}$

After the equivalence point, you have neutralized all the $\text{NaOH}$ and have a solution of excess $\text{HCl}$.

pH at 50.1 mL

$\text{Moles HCl added" = 0.0501 color(red)(cancel(color(black)("L"))) × "0.10 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.005 01 mol}$

$\text{Excess moles HCl = (0.005 01 - 0.005 00) mol = 0.000 01 mol}$

$\text{Volume = (690 + 50.1) mL = 740.1 mL = 0.7401 L}$

$\text{[HCl]" = "0.000 01 mol"/"0.7401 L" = "0.000 014 mol/L}$

$\text{pH = -log(0.000 013) = 4.87}$

pH at 51 mL

$\text{Moles HCl added" = 0.051 color(red)(cancel(color(black)("L"))) × "0.10 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.0051 mol}$

$\text{Excess moles HCl = (0.0051 - 0.005 00) mol = 0.0001 mol}$

$\text{Volume = (690 + 51) mL = 741 mL = 0.741 L}$

$\text{[HCl]" = "0.0001 mol"/"0.741 L" = "0.000 13 mol/L}$

$\text{pH = -log(0.000 13) = 3.87}$

pH at 75 mL

$\text{Moles HCl added" = 0.0750 color(red)(cancel(color(black)("L"))) × "0.10 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.007 50 mol}$

$\text{Excess moles HCl = (0.007 50 - 0.005 00) mol = 0.002 50 mol}$

$\text{Volume = (690 + 75) mL = 765 mL = 0.765 L}$

$\text{[HCl]" = "0.002 50 mol"/"0.741 L" = "0.003 27 mol/L}$

$\text{pH = -log(0.00 327) = 2.49}$

Plot pH vs Volume

Plot the points on a graph and draw a smooth curve between them.

I added a few extra points to get a smoother curve.

Your graph should something like this. 