calculate the volume(ml) of a 0.215M KOH solution that will completely neutralize each of the following?

Question

a) 2.50ml of a 0.825M H2SO4 solution
b) 18.5ml of a 0.560 M HNO3 solution
c) 5.00mL of a 3.18 M H2SO4 solution

Answer 1

All of these are done the same way.
1. Write and balance the equation.
2KOH + H2SO4 ==> K2SO4

2. Calculate moles. moles = M x L.
moles H2SO4 = M x L = 0.825 x 0.0025 L = 0.00206

3. Using the coefficients in the balanced equation, convert moles H2SO4 to moles KOH.
0.00206 moles H2SO4 x (2 moles KOH/1 mol H2SO4) = 0.00206 x 2 = 0.00413 moles KOH.

4. Then M KOH = moles KOH/L KOH. We know M and moles, solve for L.
0.215M = 0.00413/L
L = 0.00413/0.215 = 0.01919 L or 19.19 mL which rounds to 19.2 mL to three significant figures.

Answer 2

Hi, I happen to have the same problem in my chemistry book and know an easier way! My professor taught it to us, but I forgot and started searching Google and then after reading DrBob's solution realized I knew how to do it an easier way. Use the equation NMV = NMV. One side of it is the acid and other is of the base. N = the # of H+ or OH- ions. M = molarity. And V = volume.

So:
2 x 0.825 x 2.50 = 1 x 0.215 x V
V of the base = (2 x 0.825 x 2.50) / (1 x 0.215)
V of the base = 19.2 mL