Chemistry Wizard pt. 3?

[? Only the gud dye yung]

Calculate the volume in (mL) of each of the following acid solutions that would be required to neutralize 36.2 mL of 0.259 M NaOH solution.

a. 0.271 M HCI
b. 0.119 M H2SO4
c. 0.171 M H3PO4

[cstspeedy]

0.0362L of 0.259M NaOH = 0.00938 moles of OH-

a) This requires 1 mole of HCl per 1 mole OH-

0.00938 / 0.271 = 0.0346L = 34.6mL HCl

b) This has 2 moles of H+ for every mole of acid

0.119M H2SO4 = 0.238M H+

0.00938 / 0.238 = 0.0394L = 39.4mL H2SO4

c) This has 3 moles H= for every mole of acid

0.171M H3PO4 = 0.513M H+

0.00938 / 0.513 = 0.0179L = 17.9mL of H3PO4