Question

caluclate the pH in the titration of 50.00 mL of 0.200 M HNO3 by 0.100 M NaOH after the addition to the acid of solutions of

(a) 0 mL NaOH

(b) 10.00 ml NaOH

(c) 100.00 ml Na OH

(d) 150 mL NaOH

Answer 1

Ans a) The reaction between HNO₃and NaOH will be

HNO₃ + NaOH ->NaNO₃ + H₂O

Here we can see the formation of water therefore we ca take H⁺ from HNO₃and OH^- from NaOH to form H₂O

H⁺ + OH^- ->H₂O

Given

Volume of HNO₃ = 50ml =0.05L

moles of H⁺ present in the reaction = 0.05x0.2 =0.01mol H⁺

Volume of NaOH added = 0 ml

moles of OH^- added = 0

Therefore moles of H⁺ remaining in the reaction = (0.01-0) =0.01mol

Total volume of reaction=50ml+0ml

                                 =50ml =0.05L

Concentration [H⁺] = number of moles /volume

                             = 0.01/0.05

                             =0.2mol/L

pH= - log[H⁺]

    = - log(0.2)

     = 0.698

Ans b) Here

Volume of HNO₃ = 50ml = 0.05L

moles of H⁺ present in the reaction = 0.05x0.2

                                                    = 0.01molH⁺

Volume of NaOH added = 10ml=0.01L

concentration of NaOH = 0.1M

Moles of OH^- added = 0.01x0.1

                               = 0.001mol

Therefore moles of H⁺ remaining in the reaction =(0.01-0.001)

                                                                     = 0.009mol

Total volume of the solution = 50ml+10ml =60ml = 0.06L

Concentration [H⁺]=Number of moles/volume

                            =0.009/0.06

                            =0.15

pH= - log[H⁺]

    = - log[0.15]

    = 0.823

Ans c) Here

Given volume of HNO₃ = 50ml=0.05L

moles of H⁺ present = 0.05x0.2

                              = 0.01mol H⁺

Volume of NaOH added = 100ml =0.1L

moles of OH^- added = 0.1x0.1

                             = 0.01mol

Therefore moles of H⁺ remaining = (0.01-0.01)

                                                = 0

Total volume = 50+100 = 150ml =0.15L

Concentration[H⁺] = 0

pH = - log[H⁺]

    = - log(0)

     = 0

Ans d) Given volume of HNO₃ = 50ml=0.05L

moles of H⁺ present in the reaction = 0.05x0.2

                                                     = 0.01mol

Volume of NaOH added = 150ml=0.15L

moles of OH^- added = 0.15x0.1

                              = 0.015mol

Therefore moles of H⁺ remaining = (0.015-0.01)

                                                  = 0.005mol

Total volume =50+150= 200ml = 0.2L

Concentartion[H⁺]= 0.005/0.2

                          = 0.025

pH = -log(H⁺)

       = - log(0.025)

       = 1.6