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Terdapat 200 mL larutan  0,2 M, jika kedalamnya ditambah dengan 3,7 gr  maka pH campurannya adalah....

Pertanyaan

Terdapat 200 mL larutan begin mathsize 14px style Na O H end style 0,2 M, jika kedalamnya ditambah dengan 3,7 gr begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 end style maka pH campurannya adalah....

Pembahasan Soal:

Volume = 200 mL = 0,2 L

Untuk menentukan pH campuran, perlu ditentukan konsentrasi campuran basanya seperti berikut.

  • Konsentrasi begin mathsize 14px style O H to the power of minus sign end style pada begin mathsize 14px style Na O H end style 

           M subscript O H to the power of minus sign end subscript double bond M subscript Na O H space end subscript cross times space valensi

  • Mol begin mathsize 14px style O H to the power of minus sign end style pada begin mathsize 14px style Na O H end style 

            Mol space Na O H space equals space M space cross times space V space space space space space space space space space space space space space space space space space space space equals M subscript Na O H space end subscript cross times space valensi space cross times space V space space space space space space space space space space space space space space space space space space equals space 0 comma 2 space M space cross times 1 space cross times space 0 comma 2 space L space space space space space space space space space space space space space space space space space space space equals space 0 comma 04 space mol 

            

  • Mol begin mathsize 14px style O H to the power of minus sign end style pada begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 end style

           begin mathsize 14px style Mr space Ca open parentheses O H close parentheses subscript 2 space equals space Ar space Ca space plus space 2 space cross times space Ar space O space plus space 2 space cross times space Ar space H space space space space space space space space space space space space space space space space space space space space equals space 40 space plus space 2 left parenthesis 16 right parenthesis space plus space 2 left parenthesis 1 right parenthesis space space space space space space space space space space space space space space space space space space space space equals space 74 end style 

           Error converting from MathML to accessible text.

             

  • Konsentrasi begin mathsize 14px style O H to the power of minus sign end style pada begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 end style

            n subscript O H to the power of minus sign end subscript equals 2 over 1 cross times n subscript Ca open parentheses O H close parentheses subscript 2 end subscript n subscript O H to the power of minus sign end subscript equals 2 over 1 cross times 0 comma 05 n subscript O H to the power of minus sign end subscript equals 0 comma 1 space mol

 

Konsentrasi begin mathsize 14px style O H to the power of minus sign end style campuran

          Error converting from MathML to accessible text. 

 pH campurannya yaitu :

Error converting from MathML to accessible text.      

   Error converting from MathML to accessible text.

 Jadi, pH campuran antara begin mathsize 14px style Na O H end style dengan Error converting from MathML to accessible text. adalah 13 + log 7. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 01 Mei 2021

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Pertanyaan yang serupa

Tentukan pH larutan yang mengandung ion hidroksida dalam air.

Pembahasan Soal:

Diketahui : begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets space equals space 3 comma 98 cross times 10 to the power of negative sign 9 end exponent end style M

Ditanya : pH?

Jawab :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pOH space end cell equals cell space minus sign space log space open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell space minus sign space log space 3 comma 98 cross times 10 to the power of negative sign 9 end exponent end cell row pOH equals cell space 9 space minus sign space log space 3 comma 98 end cell row blank blank blank row cell pH space end cell equals cell space 14 space minus sign space pOH end cell row blank equals cell space 14 minus sign space left parenthesis 9 minus sign log space 3 comma 98 right parenthesis end cell row blank equals cell space 14 minus sign 9 space plus space log space 3 comma 98 end cell row cell pH space end cell equals cell space 5 space plus space log space 3 comma 98 end cell end table end style 

Jadi, pH larutan tersebut adalah begin mathsize 14px style 5 space plus space log space 3 comma 98 end style.

0

Roboguru

Tentukan konsentrasi ion  dalam larutan  0,01 M dan dalam larutan  0,002 M!

Pembahasan Soal:

Senyawa begin mathsize 14px style Na O H end style dan begin mathsize 14px style Ba open parentheses O H close parentheses subscript 2 end style merupakan senyawa basa kuat yang akan terionisasi sempurna dalam larutannya. Larutan begin mathsize 14px style Na O H end style 0,01 M akan terionisasi menurut persamaan reaksi berikut:undefined


begin mathsize 14px style Na O H yields Na to the power of plus sign and O H to the power of minus sign end style 
 

Berdasarkan persamaan reaksi di atas, senyawa begin mathsize 14px style Na O H end style terionisasi dan menghasilkan 1 ion begin mathsize 14px style O H to the power of minus sign end style dalam larutannya, maka valensi basanya adalah 1. Konsentrasi ion begin mathsize 14px style O H to the power of minus sign end style dapat dihitung dengan cara seperti berikut.undefined


begin mathsize 14px style open square brackets O H close square brackets to the power of minus sign equals M cross times valensi space basa open square brackets O H close square brackets to the power of minus sign equals 0 comma 01 cross times 1 open square brackets O H close square brackets to the power of minus sign equals 0 comma 01 open square brackets O H close square brackets to the power of minus sign equals 1 middle dot 10 to the power of negative sign 2 end exponent end style 


Larutan begin mathsize 14px style Ba open parentheses O H close parentheses subscript 2 end style 0,002 M akan terionisasi menurut reaksi berikut:


begin mathsize 14px style Ba open parentheses O H close parentheses subscript 2 yields Ba to the power of 2 plus sign and 2 O H to the power of minus sign end style 


Berdasarkan persamaan reaksi di atas, senyawa undefined terionisasi dan menghasilkan 2 ion begin mathsize 14px style O H to the power of minus sign end style dalam larutannya, maka valensi basanya adalah 2. Konsentrasi ion begin mathsize 14px style O H to the power of minus sign end style dapat dihitung dengan cara seperti berikut.undefined


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H close square brackets to the power of minus sign end cell equals cell M cross times valensi space basa end cell row cell open square brackets O H close square brackets to the power of minus sign end cell equals cell 0 comma 002 cross times 2 end cell row cell open square brackets O H close square brackets to the power of minus sign end cell equals cell 0 comma 004 end cell row cell open square brackets O H close square brackets to the power of minus sign end cell equals cell 4 cross times 10 to the power of negative sign 3 end exponent end cell end table end style 


Jadi, jawaban yang tepat adalah seperti pada penjelasan di atas.undefined

0

Roboguru

Berapakah dalam larutan yang mempunyai pH=12

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH and pOH end cell equals 14 row pOH equals cell 14 minus sign 12 end cell row blank equals 2 row blank blank blank row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row 2 equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 2 end exponent end cell end table end style  


Jadi, begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets end style dalam larutan yang memiliki pH 12 adalah begin mathsize 14px style 10 to the power of negative sign 2 end exponent end style.

0

Roboguru

Ketika ion hidroksida konsentrasinya  maka pH larutan sebesar ….

Pembahasan Soal:

Ion hidroksida adalah ion undefined sehingga Error converting from MathML to accessible text. yang pada akhirnya

begin mathsize 12px style open square brackets straight H to the power of plus close square brackets open square brackets OH to the power of minus close square brackets equals 10 to the power of negative 14 end exponent open square brackets straight H to the power of plus close square brackets open square brackets 10 to the power of negative 2 end exponent close square brackets equals 10 to the power of negative 14 end exponent open square brackets straight H to the power of plus close square brackets equals 10 to the power of negative 14 end exponent over 10 to the power of negative 2 end exponent equals 10 to the power of negative 12 end exponent pH equals negative log open square brackets straight H to the power of plus close square brackets pH equals negative log open square brackets 10 to the power of negative 12 end exponent close square brackets pH equals 12 end style 

0

Roboguru

Sebanyak 50 mL larutan  0,02 M dicampur dengan 50 mL larutan  0,1 M. pH campuran larutan tersebut adalah ....

Pembahasan Soal:

Larutan begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 end style merupakan basa kuat, sementara begin mathsize 14px style N H subscript 4 O H end style merupakan basa lemah. Besarnya pH campuran kedua larutan tersebut, dapat ditentukan dengan cara sebagai berikut.

1. Tentukan konsentrasi ion O H to the power of minus sign dari begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 end style


table attributes columnalign right center left columnspacing 0px end attributes row cell Ca open parentheses O H close parentheses subscript 2 space end cell rightwards arrow cell space Ca to the power of 2 plus sign space plus space 2 O H to the power of minus sign end cell row blank blank blank row cell open square brackets O H minus sign close square brackets end cell equals cell M cross times valensi end cell row blank equals cell 0 comma 02 space M cross times 2 end cell row blank equals cell 0 comma 04 space M end cell end table


2. Tentukan konsentrasi ion O H to the power of minus sign dari begin mathsize 14px style N H subscript 4 O H end style


table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript b cross times M end root end cell row blank equals cell square root of 10 to the power of negative sign 5 end exponent cross times 0 comma 1 end root end cell row blank equals cell square root of 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 10 to the power of negative sign 3 end exponent end cell row blank equals cell 0 comma 001 end cell end table


3. Tentukan konsentrasi O H to the power of minus sign campuran


table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets subscript campuran end cell equals cell fraction numerator left parenthesis open square brackets O H to the power of minus sign close square brackets cross times V right parenthesis subscript Ca open parentheses O H close parentheses subscript 2 end subscript space plus space left parenthesis open square brackets O H to the power of minus sign close square brackets cross times V right parenthesis subscript N H subscript 4 O H end subscript over denominator V subscript Ca open parentheses O H close parentheses subscript 2 end subscript space plus space V subscript N H subscript 4 O H end subscript end fraction end cell row blank equals cell fraction numerator left parenthesis 0 comma 04 space M cross times 50 space mL right parenthesis space plus space left parenthesis 0 comma 001 space M cross times 50 space mL right parenthesis over denominator left parenthesis 50 plus 50 right parenthesis space mL end fraction end cell row blank equals cell fraction numerator left parenthesis 2 plus 0 comma 05 right parenthesis space mmol over denominator 100 space mL end fraction end cell row blank equals cell fraction numerator 2 comma 05 space mmol over denominator 100 space mL end fraction end cell row blank equals cell 2 comma 05 cross times 10 to the power of negative sign 2 end exponent space M end cell end table


4. Tentukan pOH dan pH


table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log left parenthesis 2 comma 05 cross times 10 to the power of negative sign 2 end exponent right parenthesis end cell row blank equals cell 2 minus sign log space 2 comma 05 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign left parenthesis 2 minus sign log space 2 comma 05 right parenthesis end cell row blank equals cell 12 plus log space 2 comma 05 end cell end table


Dengan demikian, pH campuran larutan tersebut adalah 12 + log 2,05.

Jadi, jawaban yang benar adalah D.space

0

Roboguru

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