Answer :
[tex]\\ \tt\leadsto -log[H^+]=5.31[/tex]
[tex]\\ \tt\leadsto log[H^+]=-5.31[/tex]
[tex]\\ \tt\leadsto [H^+]=antilog(-5.31)[/tex]
[tex]\\ \tt\leadsto [H^+]=4.89\times 10^{-6}M[/tex]
[tex]\\ \tt\leadsto -log[H^+]=5.31[/tex]
[tex]\\ \tt\leadsto log[H^+]=-5.31[/tex]
[tex]\\ \tt\leadsto [H^+]=antilog(-5.31)[/tex]
[tex]\\ \tt\leadsto [H^+]=4.89\times 10^{-6}M[/tex]