$0.20\mathrm{M}$ HA ($K_a=4.0\times 10^{-5}$) is made $0.10\mathrm{M}$ in $\mathrm{NaA}$ by dissolving the solid (no volume change). Taking $[\mathrm{HA}]\approx 0.20$ and $[\mathrm{A}{}^{-}]\approx 0.10\mathrm{M}$, which statements are correct? ($\log 8=0.903$)

1. AThe degree of ionisation of HA falls to about $4.0\times 10^{-4}$.
2. BThe pH becomes $4.10$.
3. C$[\mathrm{H}{}^{+}]$ is reduced by a factor of about $35$.
4. D$K_a$ of HA decreases.