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:

molarity (M)=molL

moles HCl=0.340 mol HClL 34.0 mL x 10-3 L1 mL

moles HCl = 0.01156 mol HCl

moles NaOH=0.340 mol NaOHL44.0 mL x 10-3 L1 mL

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

molarity (M)=molL


final volume of solution=volume HCl+volumeNaOHfinal volume of solution=34.0 mL x 10-3 L1 mL+44.0 mL x 10-3 L1 mL

Final Volume of solution = 0.078 L


molarity=0.0034 mol NaOH0.078 Lmolarity=0.0436 mol NaOHL

molarity = 0.0436 M NaOH

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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 =

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