# Problem: Calculate the pH of the resulting solution if 34.0mL of 0.340 M HCl(aq) is added to(a) 44.0mL of 0.340 M NaOH(aq).pH =(b) 24.0mL of 0.440 M NaOH(aq)pH =

###### FREE Expert Solution

▪ HCl → strong binary acid
▪ NaOH → (OH- with Group 1A ion) → strong base
▪ the reaction between a strong base and strong acid → no need to create an ICE chart

Balanced reaction:              HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

(a) 34.0mL of 0.340 M HCl(aq) + 44.0mL of 0.340 M NaOH(aq)

Step 1: Calculate the initial amount (in moles) of the HCl and NaOH

Recall:

moles HCl = 0.01156 mol HCl

moles NaOH = 0.01496 mol NaOH

The amount of NaOH is greater than HCl, therefore there will be NaOH left after the reaction

Step 2: Calculate the amount of HCl reacted

Balanced reaction:                 HCl + NaOH → H2O + NaCl

1 mole of HCl reacts with 1 mole of NaOH:
mol of HCl reacted = mol NaOH reacted

moles NaOH reacted = 0.01156 mol NaOH

Step 3: Calculate the amount of NaOH left after the reaction

moles NaOH left = moles NaOH (initial) – moles NaOH (reacted)

moles NaOH left = 0.01496 mol HCl – 0.01156 mol HCl

moles NaOH left = 0.0034 mol

Step 4: Calculate the concentration of NaOH left in the solution

Final Volume of solution = 0.078 L

molarity = 0.0436 M NaOH ###### Problem Details

Calculate the pH of the resulting solution if 34.0mL of 0.340 M HCl(aq) is added to

(a) 44.0mL of 0.340 M NaOH(aq).

pH =

(b) 24.0mL of 0.440 M NaOH(aq)

pH =