Question

Part A

42.0 mL of 0.150 M NaOH and 37.6 mL of 0.370 M NaOH

Enter your answers numerically separated by a comma.

Part B

44.0 mL of 0.100 M Na2SO4 and 25.0 mL of 0.170 M KCl

Enter your answers numerically separated by commas.

Part C

3.70 g KCl in 75.0 mL of 0.280 M CaCl2 solution. Assume that the volumes are additive.

Enter your answers numerically separated by commas.

Answer 1

A)

mixed molarity = (M1 V1 + M2V2 )/ V1+V2

                        = (0.150 x 42 + 0.370 x 37.6 ) / (42+37.6)

                        = 0.254 M

NaOH = 0.254M

[Na+ ]= 0.254M

[OH-]= 0.254M

Part B

[Na+] = 2 x 44 x 0.1 / 44 + 25

            = 0.1275 M

{SO4-2] = 44 x 0.1 / 44 + 25 = 0.063 M

[K+] =25 x 0.17 / 44 + 25 = 0.062 M

[Cl-] = 25 x 0.17 / 44 + 25 = 0.062 M

Part C

KCl moles = 3.70 / 74.55 = 0.0493

KCl molarity = 0.0493 / 0.075 = 0.66M

[K+] = 0.66 M

[Ca+2] = 0.280 M

[Cl-] = 0.66 + 2 x 0.280 = 1.22 M