Posted by **Connie** on Thursday, April 28, 2011 at 11:37am.

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

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

- chemistry -
**DrBob222**, Thursday, April 28, 2011 at 1:10pm
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.

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