May I please get help on this?

Question

Calculate the pH of the resulting solution if 28.0 mL of 0.280 M HCl(aq) is added to 38.0 mL of 0.280 M NaOH(aq).

and

Calculate the pH of the resulting solution if 28.0 mL of 0.280 M HCl(aq) is added to 18.0 mL of 0.380 M NaOH(aq).

Answer 1

First, write the balanced equation:

HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l)

Next, calculate moles of each that are present:

moles HCl = 28.0 ml x 1 L / 1000 ml x 0.280 mol / L = 0.00784 moles

moles NaOH = 38.0 ml x 1 L / 1000 ml x 0.280 mol / L = 0.01064 moles

Excess moles of NaOH = 0.01064 mols - 0.00784 mols = 0.00280 moles NaOH

You could have done that with an ICE (BCA) table if you are familiar with those:

HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l)

.00784....0.01064................0...............0.............Initial

-0.00784..-0.00784..........+0.00784..+0.00784....Change

0..............0.00280..............0.00784...0.00784....Equilbrium

Total volume = 28.0 mls + 38.0 mls = 66 mls = 0.066 L

Final [NaOH] = 0.00280 mols / 0.066 L = 0.0424 M

pOH = -log 0.0424 = 1.37

pH = 14 - pOH = 14 - 1.37

pH = 12.6

For the 2nd problem, it will be the same as above except this time the HCl will be in excess because there are more moles HCl than of NaOH.

HCl + NaOH ==> NaCl + H2O

Initial moles HCl = 28.0 ml x 1 L / 1000 ml x 0.280 mol/L = 0.00784 mols

Initial moles NaOH = 18.0 ml x 1 L / 1000 ml x 0.380 mol / L = 0.00684 mols

Excess moles HCl = 0.00784 - 0.00684 = 0.001 moles HCl

Final volume = 28.0 ml + 18.0 ml = 46 ml = 0.046 L

Final [HCl] = 0.001 mols / 0.046 L = 0.0217 M

pH = -log [H⁺] = -log 0.0217

pH = 1.66