Question

a 40.0 ml sample of ca(oh)2 was titrated with 0.0704 m hcl. the volume of hcl needed to reach equivalence point was 18.2 ml. what was the concentration of ca2+ present in the original solution?4

Answer 1

Write the balanced reaction:
Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

Then we get the moles of HCl by multiplying the given concentration by the volume. I'm assuming that the given concentration, 0.0704 m, is the molarity:
M = n/V
n = M*V
n = 0.0704 M * 18.2 mL
n = 1.28128 mmol HCl

Then from the balanced reaction, we get a mole ratio of Ca(OH)2 and HCl. For every mole of Ca(OH)2 reacted, two moles of HCl are needed, so
1 mmol Ca(OH)2 / 2 mmol HCl

We multiply this ratio by the moles of HCl we got:
1.28128 mmol HCl * 1 mmol Ca(OH)2 / 2 mmol HCl = 0.64064 mmol Ca(OH)2

Finally, divide this by the given volume of Ca(OH)2:
0.64064 mmol Ca(OH)2 / 40 mL Ca(OH)2 = ?

Units in mol/L (M).