Question

what volume of 0.500 m koh is needed to completely neutralize 125.0 ml of a 0.750 m h₂so₄ solution?
h₂so₄(aq) + 2 koh(aq) → k₂so₄(aq) + 2 h₂o(l)
○ 375 ml
○ 125 ml
○ 250. ml
○ 188 ml

Explanation:

Step1: Recall the neutralization formula

For acid - base neutralization, the formula is \(n_1M_1V_1=n_2M_2V_2\), where \(n_1\) and \(n_2\) are the number of moles of \(H^+\) and \(OH^-\) ions donated per formula unit of acid and base respectively, \(M_1\) and \(M_2\) are the molarities, and \(V_1\) and \(V_2\) are the volumes.

For \(H_2SO_4\), the reaction is \(H_2SO_4 + 2KOH
ightarrow K_2SO_4+2H_2O\). So, \(n_1 = 2\) (since \(H_2SO_4\) donates 2 \(H^+\) ions), \(M_1=0.750\ M\), \(V_1 = 125.0\ mL\). For \(KOH\), \(n_2 = 1\) (since \(KOH\) donates 1 \(OH^-\) ion), \(M_2 = 0.500\ M\), and \(V_2\) is the volume we need to find.

Step2: Substitute the values into the formula

Substitute the values into \(n_1M_1V_1=n_2M_2V_2\). We get \(2\times0.750\ M\times125.0\ mL=1\times0.500\ M\times V_2\)

First, calculate the left - hand side: \(2\times0.750\times125.0=2\times93.75 = 187.5\)

Then, solve for \(V_2\): \(V_2=\frac{2\times0.750\times125.0}{0.500}\)

\(V_2=\frac{187.5}{0.500}=375\ mL\)

Answer:

375 mL