Question

In a titration 32.5 mL of 1.0 M sulfuric acid is required to neutralize 45.0 mL of sodium hydroxide. What is the concentration of the base?

Answer 1

`\large \boxed{\text{1.4 mol/L}}`

Step-by-step explanation:

(a) Balanced equation

2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

(b) Moles of H₂SO₄

`\text{Moles of H$_(2)$SO}_(4) = \text{32.5 mL H$_(2)$SO}_(4) * \frac{\text{1.0 mmol H$_(2)$SO$_(4)$}}{\text{1 mL H$_(2)$SO$_(4)$}}\\= \text{32.5 mmol H$_(2)$SO}_(4)`

(c) Moles of NaOH

The molar ratio is 2 mol NaOH:1 mol H₂SO₄.

`\text{Moles of NaOH} = \text{32.5 mmol H$_(2)$SO}_(4) *\frac{\text{2 mmol NaOH}}{\text{1 mmol H$_(2)$SO}_(4)}\\\\= \text{65 mmol NaOH}`

(d) Molar concentration of NaOH

`c = \frac{\text{moles of solute}}{\text{litres of solution}}\\\\c = (n)/(V)\\\\c= \frac{\text{65 mmol}}{\text{45.0 mL}} = \text{1.4 mol$\cdot$L$^(-1)$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{1.4 mol/L}}$}`

Note: The answer can have only two significant figures because that is all you gave for the molar concentration of the sulfuric acid.