Penjelasan:

Langkah pertama dalam pengerjaan soal ini adalah menghitung menggunakan rumus molaritas, yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space H subscript 2 S O subscript 4 end cell equals cell m over M subscript r cross times 1000 over V end cell row blank equals cell fraction numerator 4 comma 9 over denominator 98 end fraction cross times 1000 over 200 end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 1 end exponent space M end cell end table end style

Kemudian kita masukan molaritas tersebut ke rumus pH asam kuat, yaitu:

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals M cross times valensi space asam space space space space space space space space equals 2 comma 5 cross times 10 to the power of negative sign 1 end exponent cross times 2 space space space space space space space space equals 5 cross times 10 to the power of negative sign 1 end exponent pH equals minus sign log space open square brackets H to the power of plus sign close square brackets space space space space space equals minus sign log space 5 cross times 10 to the power of negative sign 1 end exponent space space space space space equals 1 minus sign log space 5 end style