Question

If you mix 25mL of 0.25 Sodium phosphate dibasic (Na₂HPO₄) with 50mL of 0.1M Sodium phosphate monobasic (NaH₂PO₄) what will be the resulting pH?

Answer 1

Number of moles of Sodium phosphate dibasic (Na₂HPO₄) , n = molarity x volume in L

                                                                                                   = 0.25 M x (25/1000) L

                                                                                                   = 0.00625 mol

Since it is dibasic it produces 2 moles of H⁺

So the number of moles of H⁺ = 2 x 0.00625 mol = 0.0125 mol

Number of moles of 0.1M Sodium phosphate monobasic (NaH₂PO₄) , n ' = molarity x volume in L

                                                                                                  = 0.1 M x (50/1000) L

                                                                                                  = 0.005 mol

Since Sodium phosphate is monobasic it produces one mole of H⁺

So number of moles of H⁺ = 0.005 mol

Upon mixing these two solutions the total number of H⁺ in the resulting solution is

                   = sum of these two solutions H⁺ ion moles

                  = 0.0125 + 0.005 mol

                  = 0.0175 mol

So the concentration of H⁺ ion in the mixture is = total number of moles / total volume in L

                                                                            = 0.0175 mol / [( 25+50) / 1000 ] L

                                                                           = 0.233 M

We know that pH = - log [H⁺]

                            = - log 0.233

                            = 0.63

Therefore the pH of the resulting solution is 0.63