(CH3)2NH2+of PKb=10.77 and acetate ion of PKb =12.04
Kb of acetate ion can be calculated by
[tex]\begin{align} K_{b} &= \frac{K_{w}}{K_{a}}\\ &= \frac{1\times 10^{-14}}{1\times10^{-2}} \\ &= 9.09 \times 10^{-13} \end{align}[/tex][tex]Kb=\frac{Kw}{Ka} \\[/tex]
=1×10^−2/1×10^−14
=9.09x 10^-13
we have the Kb value, we can now compute the pKb value using the formula below
PKb= -logKb
PKb=-log(9.09x 10^-13)
PKb=12.04
The Kb of (CH3)2NH2+ can be calculated as
[tex]\begin{align} K_{b} &= \frac{K_{w}}{K_{a}}\\ &= \frac{1\times 10^{-14}}{1\times10^{-2}} \\ &= 9.09 \times 10^{-13} \end{align}[/tex][tex]Kb=\frac{Kw}{Ka} \\[/tex]
=1×10^−14/5.9x10^−4
=1.7x10^-11
we have the Kb value, we can now compute the pKb value using the formula below
PKb= -logKb
PKb=-log(1.7x 10^-11)
PKb=10.77
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