Exercise 7 If 60.2 mL of 0427 M KOH solution are required to neutralize 10.1 mL of H_(2)SO_(4) solution, what is the concentration of the H_(2)SO_(4) solution in molarity?

To find the concentration of the $H_{2}SO_{4}$ solution in molarity, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between KOH and $H_{2}SO_{4}$.<br /><br />The balanced chemical equation for the neutralization reaction is:<br /><br />$2KOH + H_{2}SO_{4} \rightarrow K_{2}SO_{4} + 2H_{2}O$<br /><br />From the balanced equation, we can see that 2 moles of KOH react with 1 mole of $H_{2}SO_{4}$.<br /><br />Given information:<br />- Volume of KOH solution: 60.2 mL<br />- Concentration of KOH solution: 0.427 M<br />- Volume of $H_{2}SO_{4}$ solution: 10.1 mL<br /><br />Step 1: Calculate the number of moles of KOH.<br />Number of moles of KOH = Concentration of KOH × Volume of KOH solution<br />Number of moles of KOH = 0.427 M × 0.0602 L = 0.0257 moles<br /><br />Step 2: Calculate the number of moles of $H_{2}SO_{4}$.<br />Since 2 moles of KOH react with 1 mole of $H_{2}SO_{4}$, the number of moles of $H_{2}SO_{4}$ is half the number of moles of KOH.<br />Number of moles of $H_{2}SO_{4}$ = 0.0257 moles / 2 = 0.01285 moles<br /><br />Step 3: Calculate the concentration of the $H_{2}SO_{4}$ solution.<br />Concentration of $H_{2}SO_{4}$ = Number of moles of $H_{2}SO_{4}$ / Volume of $H_{2}SO_{4}$ solution<br />Concentration of $H_{2}SO_{4}$ = 0.01285 moles / 0.0101 L = 1.27 M<br /><br />Therefore, the concentration of the $H_{2}SO_{4}$ solution is 1.27 M.