Question

Find pH of 0.105 M in formic acid and 5.0×10−2 M in hypochlorous acid

Answer 1

Let a be the dissociation of the weak acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids α is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

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(1) For formic acid

K_a = 1.78x10^-4

          c = concentration = 0.105 M

Plug the values we get a = 0.041

So[H+] = ca = 4.32x10^-3 M

pH = - log[H⁺]

    = - log (4.32x10^-3 )

   = 2.36

(2) For hypochlorous acid

K_a = 2.82x10^-8

          c = concentration = 5.0x10^-2 M

Plug the values we get a = 7.50x10^-4

So[H⁺] = ca = 3.75 x10^-5 M

pH = - log[H⁺]

    = - log (3.75 x10^-5 )

   = 4.42