Question

calculate the ph of 0.02 M sulfuric acid​

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

$H_2SO_4\:\longrightarrow\:2H^+\:+\:(SO_4)^2-\\\\[H^+]\:=\:2×0.02\:=\:0.04\: M$

$P^H\:=\:-\:log\:[H^+]\\=\:-\:log(0.04)\\\:\:\:\:=\:1.4$

$\huge \green{ \underbrace {\bigcirc} _{ \red \odot} \: }$

Answer 2

what is cloub what si school