Question

what volume of 0.830 m koh is needed to completely neutralize 50.0 ml of a 0.500 m h₂so₄ solution?
h₂so₄(aq) + 2 koh(aq) → k₂so₄(aq) + 2 h₂o(l)
○ 100. ml
○ 30.1 ml
○ 120. ml
○ 60.2 ml

Explanation:

Step1: Calculate moles of \(H_2SO_4\)

Use the formula \(n = C\times V\). Given \(C = 0.500\ M\) and \(V=50.0\ mL=0.0500\ L\).
\(n_{H_2SO_4}=0.500\ M\times0.0500\ L = 0.0250\ mol\)

Step2: Determine moles of \(KOH\) using stoichiometry

From the balanced equation \(H_2SO_4(aq)+2KOH(aq)\to K_2SO_4(aq)+2H_2O(l)\), the mole ratio \(n_{KOH}:n_{H_2SO_4}=2:1\).
So \(n_{KOH}=2\times n_{H_2SO_4}=2\times0.0250\ mol = 0.0500\ mol\)

Step3: Calculate volume of \(KOH\) solution

Use the formula \(V=\frac{n}{C}\). Given \(n = 0.0500\ mol\) and \(C = 0.830\ M\).
\(V_{KOH}=\frac{0.0500\ mol}{0.830\ M}\approx0.0602\ L = 60.2\ mL\)

Answer:

60.2 mL