No 17 1. How many milliliters of 0.10 m H_(2)SO_(4) must be added to 100 mL of 0.10 M NaOH to give a solution that is 0.050 M in H_(2)SO_(4) ? Assume volumes are additive. A samnle of nure KHC_(2)O_(4)cdot H_(2)C_(2)O_(4)cdot 2H_(2)O

To solve this problem, we need to find the volume of 0.10 M $H_{2}SO_{4}$ that needs to be added to 100 mL of 0.10 M NaOH to achieve a final concentration of 0.050 M in $H_{2}SO_{4}$.<br /><br />Given information:<br />- Initial concentration of $H_{2}SO_{4}$: 0.10 M<br />- Initial volume of NaOH solution: 100 mL<br />- Final concentration of $H_{2}SO_{4}$: 0.050 M<br /><br />Let's denote the volume of 0.10 M $H_{2}SO_{4}$ to be added as V mL.<br /><br />The total volume of the final solution will be (100 + V) mL.<br /><br />The amount of $H_{2}SO_{4}$ in the final solution can be calculated using the formula:<br /><br />Amount of $H_{2}SO_{4}$ = Concentration × Volume<br /><br />The amount of $H_{2}SO_{4}$ in the initial 100 mL of 0.10 M NaOH solution is:<br /><br />Amount of $H_{2}SO_{4}$ in NaOH solution = 0.10 M × 0.100 L = 0.010 mol<br /><br />The amount of $H_{2}SO_{4}$ in the final solution should be 0.050 M × (100 + V) mL = 0.050 M × (0.100 + V/1000) L = 0.050 × (0.100 + V/1000) mol<br /><br />Since the amount of $H_{2}SO_{4}$ in the final solution is the sum of the amount of $H_{2}SO_{4}$ in the initial NaOH solution and the amount of $H_{2}SO_{4}$ added, we can set up the equation:<br /><br />0.010 mol + 0.10 M × V/1000 L = 0.050 × (0.100 + V/1000) mol<br /><br />Solving this equation for V, we get:<br /><br />V = 100 mL<br /><br />Therefore, 100 mL of 0.10 M $H_{2}SO_{4}$ must be added to 100 mL of 0.10 M NaOH to give a solution that is 0.050 M in $H_{2}SO_{4}$.