Question

Calculate the pH for the excess ions that will be left when 23.2mL of 1.58 mol/L HCl(aq) added 18.9mL of 3.5mol/L NaOH(aq).​

Answer 1

The pH for the excess ions that will be left when 23.2mL of 1.58 mol/L HCl(aq) is added to 18.9mL of 3.5mol/L NaOH(aq) is 13.84.

• As acid and base will give a neutralisation reaction so, the resulting pH will be determined by the left amount.
• The number of millimoles of HCl is 23.2 × 1.58 = 36.65.
• The number of millimoles of NaOH is 18.9 × 3.5 = 66.15
• The number of millimoles of base left after neutralisation is: 66.15 - 36.65 = 29.5
• Now, the concentration of left NaOH is
• $\frac{29.5}{18.9 + 23.2}= \frac{29.5}{42.1} = 0.7 \: moles \: per \: litre$
• $Now, pOH \: is -log(OH) = -log(0.7) \: = 0.155$.
• pH = 14 - pOH = 14 - 0.155 = 13.84

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