Question

Determine the pH of a solution when 21.5 mL of 0.021 M HNO3 is mixed with 17.5 mL of

A) 0.0070 M NaOH.

B) distilled water.

C) 0.0070 M HCl

D) 0.084 M KOH.

Answer 1

A)

[H+ ] = (21.5/1000)*).021 = 4.515*10^-4                                 (from 21.5 ml of 0.021MHNO3)

[OH - ] = (17.5/1000)* .0070 = 1.225*10^-4                          (from 17.5 ml of 0.0070M NaOh sol)

[H+ ] - [ OH -] = 4.515*10^-4 - 1.225*10^-4   = 3.29*10^-4

so excess [H+ ] = 3.29*10^-4

so pH = - log[ H+] = - log (3.29*10^-4   = 3.483

B)

[H+ ] = (21.5/1000)*0.021 = 4.515*10^-4             ( from 21.5 ml of 0.021 M HNO3...H2) gives no excess H+)

pH = - log[H+ ] = - log (4.515*10^-4) = +3.345

C)

[H+ ] = (21.5/1000)*0.021 + 17.5*0.0070/1000 = 5.74*10^-4         (from 21.5 ml of 0.021 M HNO3 and 17.5 ml of

                                                                                                    0.0070 M HCl sol)

pH = -log [ H+] = - log (5.74*10^-4)

                         = +3.24

D)

   [H+ ] = (21.5/1000)*0.021 =4.515*10^-4             (21.5 ml of 0.021 M of HNO3)

         [OH - ] = (17.5/1000) * 0.084 = 1.47*10^-3           ( from 17.7 mol 0.084 MKOH sol)

-[H+ ] + [OH - ] = -4.515*10^-4 + 1.47*10^-3   + 1.0185*10^-3

pOH = - log (OH-) = - log(1.085*10^-3)   = +2.992

so pH = 14 - pOH = 14 - 2.992 = 11.00796