Question

Determine the molarity of a NaOH solution when each of the following amounts of acid neutralizes 25.0 mL of the NaOH solution (a) 15.0 ml. of 0.250 M HNO3 (b) 10.0 mL of 0.500 M H SO4 (c) 20.8 ml. of 1.00 M HCI (d) 15.0 mL of 0.100 MH3PO4

Answer 1

a) HNO3 + NaOH -----------> NaNO3 + H2O

Molarity M1 = 0.25 M

Volume V1 = 15.0 ml

Number of moles n1 = 1 mole of Sodium hydroxide

Molarity M2 = x

Volume V2 = 25 ml

Number of moles n2 = 1

M1V1 / n1 = M2V2 / n2

M2 = (M1V1) * (n2 / (n1V2))

= (0.25 * 15.0) * (1 / (1 * 25))

= 0.15 M

Molarity of NaOH is 0.15 M

b)   H2SO4 + 2NaOH ---------> Na2SO4 + H2O

Molarity M2 = x

Volume V1 = 10 ml

Number of moles n2 = 2 moles

M1V1 / n1 = M2V2 / n2

M2 = M1V1 * n2 / n1V2

= (0.5 * 10) * (2 / 1 * 25)

= 0.4 M

Molarity of NaOH is 0.4 M

c)  HCl + NaOH ---------> NaCl + H2O

Molarity M1 = 1 M

Volume V1 = 20.8 ml

Number of moles n2 = 1 moles

M1V1 / n1 = M2V2 / n2

M2 = M1V1 * (n2 / n1V2)

= (1 * 20.8) * (1 / 1 * 25)

= 0.832M

Molarity of NaOH = 0.832 M

d)  H3PO4 + 3 NaOH --------> Na3PO4 + 3H2O

Molarity M1 = 0.1 M

Volume V1 = 15 ml

Number of moles n1 = 1 moles of Sodium Hydroxide

Molarity M2 = x

Volume V2 = 25 ml

Number of moles n2 = 3 moles

M1V1 / n1 = M2V2 / n2

M2 = (M1V1) * (n2 / n1V2)

= (0.1 * 15) * (3 / (1 * 25))

=  0.18 M

Molarity of NaOH is 0.18 M