If 25.0 mL of 1.28 M H2SO4 can be neutralized with 27.29mL of KOH, what is the molarity of the KOH??

Question

The equation is H2SO4+2KOH-->K2SO4+2H2O

PLease help!

Answer 1

the idea here is that at the point of neutral, the number of moles of H+ from the acid = number of moles OH- from the base.

moles of H+ (also called equivalents of H+) =
volume of acid x molarity of acid x number of H+ / molecule of acid. same for base.... so here goes

.0250 L H2SO4 x 1.28 moles H2SO4 x (2 moles H+ / 1 mole H2SO4 ) = .02729 L KOH x M moles KOH/L KOH x (1 mole OH- / 1 mole KOH)

gives

M = .0640 / .02729 moles KOH / L KOH = 2.34 molar KOH

Answer 2

0.025 L (25.0 mL) of 1.28 M H2SO4 contains 0.025x1.28 moles of H2SO4 and it contains twice that amount of moles of replaceable hydrogen ions. Moles of Hydrogen ions = 2x0.025x1.28=0.064 moles of hydrogen ions. Since 27.29 mL (0.02729 L) of KOH neutralized the 0.064 moles of H+, the 27.29 mL of KOH must have contained 0.064 moles of KOH.
0.064 moles / 0.02729 L = 2.345 moles per liter of KOH. The KOH is 2.345 M.

Answer 3

2*25*1.28/27.29 = 2.34518139 mmoles

Thus molarity = 0.0859355584