Question

What is the molar mass of the acid if a titration of 4.12 g of an acid requires 49.29mL of 0.250M NaOH to reach the equivalence point?

At a certain temperature the K_sp of Ni(OH)₂ is 6.35 x 10^-16. What is the molar soliubility of nickel(II) hydroxide in water in units of M (mol/liter)?

At a ceratin temperature the solubility of Ag₃PO₄ is 4.08 x 10^-5M. What is the K_sp at this temperature?

Answer 1

HA +     NaOH -------------> NaA + H2O

1 mole     1 mole

no of moles of NaOH = molarity * volume in L

                                  = 0.25*0.04929 = 0.0123225moles

from balanced equation

1 moles of NaOH react with 1 moles of HA

0.0123225 moles of NaOH react with 0.0123225 moles of HA

molar mass of HA = weight of acid/no of moles of acid

                             = 4.12/0.0123225 = 334.35g/mole

Ni(OH)2 ---------------> Ni⁺² (aq) + 2OH^-

                                    s                 2s

Ksp   = [Ni⁺²][OH-]²

6.35*10^-16   = s*(2s)²

6.35*10^-16   = 4s³

s³               = 6.35*10^-16/4

                   = 15.875*10^-15

s                = 2.5*10^-5 M

[Ni⁺²]       = s = 2.5*10^-5 M

[OH-]       = 2s = 5*10^-5 M

Ag3Po4 -------------> 3Ag⁺ (aq) + PO4^3-

                               3s                  s

Ksp     = [Ag⁺]³ [PO4^3-]

           = (3s)³ s

          = 27s⁴

        = 27*(4.08*10^-5 )²

         = 4.49*10^-8