Question

4.81 (a) What volume of 0.115 M HClO4 solution is needed to

neutralize 50.00 mL of 0.0875 M NaOH? (b) What volume of
0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)2? (c) If
25.8 mL of an AgNO3 solution is needed to precipitate all the
cl ions in a 785-mg sample of KCl (forming AgCl), what is
the molarity of the AgNO3 solution? (d) If 45.3 mL of a 0.108
M HCl solution is needed to neutralize a solution of KOH,
how many grams of KOH must be present in the solution?

Answer 1

To neutralize 50.00 mL of 0.0875 M NaOH, you will need 38.04 mL of 0.115 M HClO4 solution.

Step-by-step explanation:

(a) To find the volume of HClO4 solution needed to neutralize 50.00 mL of NaOH, we use the balanced equation: NaOH + HClO4 → NaClO4 + H2O From the equation, we can see that the ratio of NaOH to HClO4 is 1:1. Therefore, the number of moles of NaOH is equal to the number of moles of HClO4 needed. We can calculate the number of moles of NaOH: moles of NaOH = volume of NaOH solution (in L) × concentration of NaOH (in M) = 0.050 L × 0.0875 M = 0.004375 moles of NaOH Since the ratio of NaOH to HClO4 is 1:1, we need 0.004375 moles of HClO4. We can now find the volume: volume of HClO4 solution = moles of HClO4 / concentration of HClO4 = 0.004375 moles / 0.115 M = 0.03804 L = 38.04 mL