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Hitunglah pH larutan: CH3​COONa1M(Ka​CH3​COOH=1x10−5)

Pertanyaan

Hitunglah pH larutan:


begin mathsize 14px style C H subscript 3 C O O Na space 1 space M space left parenthesis K subscript a space C H subscript 3 C O O H equals 1 space x space 10 to the power of negative sign 5 end exponent right parenthesis end style 

  1. ... undefined 

  2. ... undefined 

Pembahasan Soal:

Diketahui:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space C H subscript 3 C O O Na end cell equals cell 1 M end cell row cell M space C H subscript 3 C O O H end cell equals cell 1 cross times 10 to the power of negative sign 5 end exponent end cell row Kw equals cell 10 to the power of negative sign 14 end exponent end cell end table end style 

Ditanya: begin mathsize 14px style pH space C H subscript 3 C O O Na end style?

Jawab:

begin mathsize 14px style C H subscript 3 C O O Na subscript begin italic style left parenthesis italic a italic q right parenthesis end style end subscript yields C H subscript 3 C O O to the power of minus sign subscript begin italic style left parenthesis italic a italic q right parenthesis end style end subscript plus Na to the power of plus sign subscript begin italic style left parenthesis italic a italic q right parenthesis end style end subscript end style 

  • mencari konsentrasi begin mathsize 14px style O H to the power of minus sign end style 

begin mathsize 14px 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 Kw over Ka n open square brackets G close square brackets end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent 1 open square brackets 1 M close square brackets end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 9 end exponent end root 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 4 comma 5 end exponent end cell end table end style 

  • mencari pOH

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log open parentheses O H to the power of minus sign close parentheses end cell row pOH equals cell negative sign log open parentheses 10 to the power of negative sign 4 comma 5 end exponent close parentheses end cell row pOH equals cell 4 comma 5 end cell end table end style 

  • mencari nilai pH

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 4 comma 5 end cell row pH equals cell 9 comma 5 end cell end table end style 

Jadi, nilai pH larutan begin mathsize 14px style C H subscript 3 C O O Na space 1 M end style adalah begin mathsize 14px style 9 comma 5 end style 

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Dijawab oleh:

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Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Hitunglah pH campuran berikut ! 20 mL HCOOH 0,3 M dengan 40 mL larutan NaOH 0,15 M (Ka = 10-4)

Pembahasan Soal:

menentukan mol masing-masing :

n space H C O O H space equals space M space X space V space space space space space space space space space space space space space space equals space 0 comma 3 space X space 20 space space space space space space space space space space space space space space equals space 6 space mmol 

n space Na O H space equals M space cross times space V space space space space space space space space space space space equals 0 comma 15 space cross times space 40 space space space space space space space space space space space equals 6 space mmol space 

menentukan  mol garam yang terbentuk :

 

menentukan konsentrasi garam yang terbentuk :

begin mathsize 11px style V space total space equals space 20 plus 40 equals 60 space ml M space garam space equals fraction numerator n space over denominator V space total end fraction M space garam space equals fraction numerator 6 space mmol over denominator 60 space ml end fraction M space garam space equals 0 comma 1 space M end style  

menentukan nilai [OH-] :

open square brackets O H to the power of minus sign close square brackets equals square root of K subscript w over K subscript a cross times M space garam end root open square brackets O H to the power of minus sign close square brackets equals square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 4 end exponent cross times left square bracket 0 comma 1 right square bracket end root open square brackets O H to the power of minus sign close square brackets equals square root of 10 to the power of negative sign 15 end exponent over 10 to the power of negative sign 4 end exponent end root open square brackets O H to the power of minus sign close square brackets equals square root of 10 to the power of negative sign 11 end exponent end root open square brackets O H to the power of minus sign close square brackets equals 10 to the power of negative sign 5 comma 5 end exponent 

menentukan pH yang terbentuk :

pOH space equals space minus sign space log space open square brackets O H to the power of minus sign close square brackets pOH space equals space minus sign space log space open square brackets 10 to the power of negative sign 5 comma 5 end exponent close square brackets pOH space equals 5 comma 5  pH space equals space 14 minus sign open parentheses pOH close parentheses pH space equals space 14 minus sign left parenthesis 5 comma 5 right parenthesis pH space equals 8 comma 5 

Jadi, pH campuran adalah 8,5 

0

Roboguru

Sebanyak 300mL larutan HCOOH0,2M (pH=4,5) dicampur dengan 200mL larutan NaOH0,3M. Berapa pH larutan setelahdicampurkan?

Pembahasan Soal:

  • Menghitung mol begin mathsize 14px style H C O O H end style

Error converting from MathML to accessible text.

 

  • Menghitung mol begin mathsize 14px style Na O H end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na O H end cell equals cell M cross times v end cell row blank equals cell 0 comma 3 space M cross times 200 space mL end cell row blank equals cell 60 space mmol end cell end table end style 
 

  • Masukkan ke tabel MRS
     


 

 

Karena pada tabel MRS tidak ada sisa pereaksi, campuran tersebut akan menghasilkan garam yang terhidrolisis sebagian karena pada garam terdapan ion H C O O to the power of minus sign yang terhidrolisis karena berasal dari ion asam lemah.

  • Ionisasi HCOONa

H C O O Na subscript space open parentheses aq close parentheses space end subscript yields H C O O to the power of minus sign subscript space open parentheses aq close parentheses space end subscript plus Na to the power of plus sign subscript space open parentheses aq close parentheses space end subscript 

  • Reaksi Hidrolisis ion H C O O to the power of minus sign

H C O O to the power of minus sign subscript open parentheses aq close parentheses space end subscript plus space H subscript 2 O subscript open parentheses l close parentheses end subscript equilibrium H C O O H subscript open parentheses aq close parentheses space end subscript and O H to the power of minus sign subscript open parentheses aq close parentheses space end subscript

  • Menghitung konsentrasi begin mathsize 14px style H C O O Na end style 

Error converting from MathML to accessible text. 
 

  • Menghitung begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 4 , 5 end cell row cell left square bracket H to the power of plus right square bracket end cell equals cell 1 cross times 10 to the power of negative sign 4 , 5 end exponent end cell end table end style
 

  • Menghitung begin mathsize 14px style K subscript a end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left square bracket H to the power of plus right square bracket end cell equals cell square root of K subscript a cross times M end root end cell row cell 1 cross times 10 to the power of negative sign 4 , 5 end exponent end cell equals cell square root of K subscript a cross times 2 cross times 10 to the power of negative sign 1 end exponent end root end cell row cell left parenthesis 1 cross times 10 to the power of negative sign 4 , 5 end exponent right parenthesis squared end cell equals cell left parenthesis square root of K subscript a cross times 2 cross times 10 to the power of negative sign 1 end exponent end root right parenthesis squared end cell row cell 1 cross times 10 to the power of negative sign 9 end exponent end cell equals cell K subscript a cross times 2 cross times 10 to the power of negative sign 1 end exponent end cell row cell K subscript a end cell equals cell fraction numerator 1 cross times 10 to the power of negative sign 9 end exponent over denominator 2 cross times 10 to the power of negative sign 1 end exponent end fraction end cell row cell K subscript a end cell equals cell 5 cross times 10 to the power of negative sign 9 end exponent end cell end table end style

 

 

 

 


 

 

 

  • Menghitung konsentrasi begin mathsize 14px style O H to the power of minus sign end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left square bracket O H to the power of minus sign right square bracket end cell equals cell square root of K subscript w over K subscript a cross times open square brackets H C O O Na close square brackets cross times valensi end root end cell row cell left square bracket O H to the power of minus sign right square bracket end cell equals cell square root of fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 5 cross times 10 to the power of negative sign 9 end exponent end fraction cross times 0 , 12 cross times 1 end root end cell row cell left square bracket O H to the power of minus sign right square bracket end cell equals cell square root of 24 cross times 10 to the power of negative sign 8 end exponent end root end cell row blank equals cell 4 , 89 cross times 10 to the power of negative sign 4 end exponent end cell end table end style

 

 

 

 

 

  • Menghitung undefined 

Error converting from MathML to accessible text.


 

  • Menghitung begin mathsize 14px style pH end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign open parentheses 4 minus sign log space 4 , 89 close parentheses end cell row blank equals cell 10 plus log space 4 , 89 end cell end table end style

 



Jadi, pH larutan setelah dicampurkan adalah begin mathsize 14px style 10 plus log space 4 comma 89 end style.

0

Roboguru

pH dari larutan garam NH4​Cl 0,2 M jika diketahui KbNH3​=2×10−5 mol/L adalah ....

Pembahasan Soal:

Senyawa N H subscript 4 Cl terbentuk dari basa lemah dan asam kuat, sehingga garamnya bersifat asam.


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 square root of Kw over Kb cross times open square brackets garam close square brackets end root end cell row blank equals cell square root of fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 2 cross times 10 to the power of negative sign 5 end exponent end fraction cross times 2 cross times 10 to the power of negative sign 1 end exponent end root end cell row blank equals cell square root of 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 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 10 to the power of negative sign 5 end exponent end cell row blank equals 5 end table 


Jadi, jawaban yang tepat adalah B.

0

Roboguru

pH larutan NaCN 0,01 M jika diketahui Ka​HCN10−10adalah ....

Pembahasan Soal:

Larutan NaCN 0,01 M merupakan garam yang bersifat basa.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Na C N end cell rightwards arrow cell Na to the power of plus sign and C N to the power of minus sign end cell row cell C N to the power of minus sign and H subscript 2 O end cell rightwards harpoon over leftwards harpoon cell H C N and O H to the power of minus sign end cell row blank blank blank 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 w over K subscript a cross times open square brackets Na C N close square brackets cross times valensi end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 10 end exponent cross times 0 comma 01 cross times 1 end root 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 3 end exponent end cell row blank blank blank row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space 10 to the power of negative sign 3 end exponent end cell row pOH equals 3 row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 3 end cell row pH equals 11 end table end style 

Jadi, jawaban yang benar adalah C.

0

Roboguru

Sebanyak 100 mL larutan CH3​COONa mempunyai pH = 10. Jika Ka​CH3​COOH=10−5 dan Mr  = 82, tentukan massa garam CH3​COONa yang terlarut dalam 100 mL larutan !

Pembahasan Soal:

Langkah 1: menghitung begin mathsize 14px style begin bold style open square brackets O H to the power of minus sign close square brackets end style end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell pKw bond pH end cell row blank equals cell 14 minus sign 10 end cell row blank equals 4 end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row 4 equals cell negative sign log space 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 4 end exponent end cell end table end style 


Langkah 2: menghitung begin mathsize 14px style begin bold style open square brackets C H subscript 3 C O O Na close square brackets end style end style 

begin mathsize 14px 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 w over K subscript a open square brackets C H subscript 3 C O O Na close square brackets end root end cell row cell 10 to the power of negative sign 4 end exponent end cell equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent open square brackets C H subscript 3 C O O Na close square brackets end root end cell row cell open parentheses 10 to the power of negative sign 4 end exponent close parentheses squared end cell equals cell open parentheses square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent open square brackets C H subscript 3 C O O Na close square brackets end root close parentheses squared end cell row cell 10 to the power of negative sign 8 end exponent end cell equals cell 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent open square brackets C H subscript 3 C O O Na close square brackets end cell row cell open square brackets C H subscript 3 C O O Na close square brackets end cell equals cell fraction numerator 10 to the power of negative sign 8 end exponent cross times 10 to the power of negative sign 5 end exponent over denominator 10 to the power of negative sign 14 end exponent end fraction end cell row blank equals cell 10 space M end cell end table end style


Langkah 3: menghitung massa begin mathsize 14px style C H subscript bold 3 C O O Na end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets C H subscript 3 C O O Na close square brackets end cell equals cell m over Mr cross times 1000 over V end cell row cell 10 space M end cell equals cell m over 82 cross times fraction numerator 1000 over denominator 100 space mL end fraction end cell row m equals cell 82 space gram end cell row blank blank blank end table end style

Jadi, massa garam begin mathsize 14px style C H subscript 3 C O O Na end style yang terlarut adalah 82 gram.

0

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