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Tentukan pH larutan jika 100 mL larutan  0,1 M direaksikan dengan 99,9 mL larutan  0,1 M (anggap volume campuran = 200 mL).

Pertanyaan

Tentukan pH larutan jika 100 mL larutan H Cl 0,1 M direaksikan dengan 99,9 mL larutan Na O H 0,1 M (anggap volume campuran = 200 mL).

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H Cl end cell equals cell M space asam cross times V space asam end cell row blank equals cell 0 comma 1 space M cross times 100 space ml end cell row blank equals cell 10 space mmol end cell end table 


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na O H end cell equals cell M thin space basa cross times V space basa end cell row blank equals cell 0 comma 1 space M cross times 99 comma 9 space ml end cell row blank equals cell 9 comma 99 space mmol end cell end table 


 


HCl tidak habis bereaksi sehingga, dapat dihitung konsentrasi sisa asam


table attributes columnalign right center left columnspacing 0px end attributes row cell M space H Cl end cell equals cell fraction numerator n space sisa over denominator V space total end fraction end cell row blank equals cell fraction numerator 0 comma 01 space mmol over denominator 200 space ml end fraction end cell row blank equals cell 0 comma 00005 space M end cell row blank equals cell 5 cross times 10 to the power of negative sign 5 space end exponent M end cell end table 


table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell alpha cross times M end cell row blank equals cell 1 cross times 5 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 cross times 10 to the power of negative sign 5 end exponent end cell row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 5 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 5 end cell end table 


Jadi, pH larutan tersebut adalah 5 - log 5.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 05 Mei 2021

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