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Sebanyak 10 mL larutan  mengandung 20% berat (mass...

Sebanyak 10 mL larutan begin mathsize 14px style Na O H end style mengandung 20% berat Na O H(massa jenis = 1,2 begin mathsize 14px style gr space mL to the power of negative sign 1 end exponent end style) diencerkan dengan penambahan air hingga volumenya 500 mL. Tentukan pH larutan setelah pengenceran!

begin mathsize 14px style open parentheses Ar colon space Na equals 23 space g space mol to the power of negative sign 1 end exponent comma space O equals 16 space g space mol to the power of negative sign 1 end exponent comma space dan space H equals 1 space g space mol to the power of negative sign 1 end exponent close parentheses end style

Jawaban:

Rumus pengenceran :

begin mathsize 14px style M subscript 1 point V subscript 1 double bond M subscript 2 point V subscript 2 end style

Keterangan :

begin mathsize 14px style M subscript 1 end style= Konsentrasi larutan sebelum ditambah air.

begin mathsize 14px style V subscript 1 end style= Volume larutan sebelum ditambah air.

begin mathsize 14px style M subscript 2 end style= Konsentrasi larutan setelah ditambah air.

begin mathsize 14px style V subscript 2 end style= Volume larutan setelah ditambah air.

Molaritas begin mathsize 14px style Na O H end style sebelum ditambah air begin mathsize 14px style open parentheses M subscript 1 close parentheses end style

begin mathsize 14px style M equals fraction numerator 10 cross times rho cross times percent sign over denominator Mr end fraction space space space equals fraction numerator 10 cross times 1 comma 2 cross times 20 over denominator 40 end fraction equals 6 end style

Molaritas  begin mathsize 14px style Na O H end style setelah ditambah air begin mathsize 14px style open parentheses M subscript 2 close parentheses end style

begin mathsize 14px style M subscript 1 cross times V subscript 1 double bond M subscript 2 cross times V subscript 2 space space space space 6 cross times 10 equals M subscript 2 cross times 500 space space space space space space space space space M subscript 2 equals 0 comma 12 end style

pH larutan  begin mathsize 14px style Na O H end style setelah diencerkan :

begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets equals M cross times b space space space space space space space space space space space equals 1 comma 2 middle dot 10 to the power of negative sign 1 end exponent cross times 1 space space space space space space space space space space space equals 1 comma 2 middle dot 10 to the power of negative sign 1 end exponent pOH equals minus sign log space open square brackets O H to the power of minus sign close square brackets space space space space space space space space equals minus sign log space open parentheses 1 comma 2 middle dot 10 to the power of negative sign 1 end exponent close parentheses space space space space space space space space equals 1 minus sign log space 1 comma 2 space pH and pOH equals 14 space space space space space space space space space space space space space pH equals 14 minus sign left parenthesis 1 minus sign log space 1 comma 2 right parenthesis space space space space space space space space space space space space space pH equals 13 plus log space 1 comma 2 end style

Maka, pH larutan undefined setelah diencerkan adalah 13+log 1,2.space

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