Question

how much is a working solution will she need to neutralize 10.5 ML of 0.250MH3 PO four

C. How much of the working solution will she need to neutralize 10.5 mL of 0.250 M H3PO4?

Please answer C and show steps.

Thank you!

Answer 1

15 g NaOH is present in 150 mL of NaOH stock solution

Hence the molarity of stock solution, M_s = (weight/Molar mass)/Volume in L

                                                                = (15g/40 gmol^-1)/0.150L

                                                                = 2.5 mol/L = 2.5 M

22.5 ml (V_s)of the stock solution is made to 250 mL solution

Hence the molarity of new solution can be calculated using the relation

M_sV_s = M_WV_W

M_W and V_W are molarity and volume of the working solution.

Hence M_w = M_sV_s / V_W

                = 2.5 M x 22.5 mL / 250 ML                              (V_W = 250 mL, given)

                = 0.225 M

Volume of working solution to neutralize the H3PO4 solution

We can use the equation

n_wM_wV_w’ = n_pM_pV_p

where n_w = valency of NaOH (working solution) = 1

n_p = valency of phosphoric acid = 3

M_p is molarity of phosphoric acid = 0.250 M

V_p is volume of phosphoric acid = 10.5 mL

Hence, the volume of the working solution,

V_w’ = n_pM_pV_p/n_wM_w

                                     = 3 x 0.250M x 10.5 mL/ (1x 0.225M)

                                     = 35 mL

Hence, the required volume of working solution for neutralization is 35 mL