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Sebanyak 100 ml larutan 0,2 M direaksikan dengan 100 ml larutan NaOH 0,2M menurut reaksi; Jika , maka pH larutan yang terbentuk adalah....

Sebanyak 100 ml larutan begin mathsize 14px style C H subscript 3 C O O H end style 0,2 M direaksikan dengan 100 ml larutan NaOH 0,2M menurut reaksi;


begin mathsize 14px style C H subscript 3 C O O H and Na O H yields C H subscript 3 C O O Na and H subscript 2 O end style 


Jika begin mathsize 14px style Ka space C H subscript 3 C O O H space equals 1.10 to the power of negative sign 5 end exponent end style, maka pH larutan yang terbentuk adalah....undefined 

  1. 2undefined 

  2. 3undefined 

  3. 5undefined 

  4. 8undefined 

  5. 9undefined 

Jawaban:

Hal pertama yang harus dilakukan adalah mencari jumlah mol dari setiap senyawa tersebut sebelum dan juga sesudah reaksi untuk mengetahui jenis reaksi yang terjadi.
 

begin mathsize 14px style Mol space C H subscript 3 C O O H double bond M cross times V Mol space C H subscript 3 C O O H equals 0 comma 2 begin inline style bevelled mol over L end style cross times 100 mL cross times 10 to the power of negative sign 3 end exponent begin inline style bevelled L over mL end style Mol space C H subscript 3 C O O H equals 0 comma 02 space mol  Mol space Na O H double bond M cross times V Mol space Na O H equals 0 comma 2 begin inline style bevelled mol over L end style cross times 100 mL cross times 10 to the power of negative sign 3 end exponent begin inline style bevelled L over mL end style Mol space Na O H equals 0 comma 02 space mol end style  
 

Reaksi asam basa:


 


Jadi setelah reaksi, terbentuk larutan garam begin mathsize 14px style C H subscript 3 C O O Na end style sebanyak 0,02 mol. Larutan garam ini akan terionisasi membentuk begin mathsize 14px style C H subscript 3 C O O to the power of minus sign end style dan begin mathsize 14px style Na to the power of plus sign end style. undefined akan menghidrolisis air sehingga terbentuk ion begin mathsize 14px style O H to the power of minus sign end style.


begin mathsize 14px style C H subscript 3 C O O Na yields C H subscript 3 C O O to the power of minus sign and Na to the power of plus sign 0 comma 02 space mol space space space space space space space space space space space 0 comma 02 space mol C H subscript 3 C O O to the power of minus sign and H subscript 2 O equilibrium C H subscript 3 C O O H and O H to the power of minus sign 0 comma 02 space mol end style  


Kemudian mengubah mol menjadi konsentrasi,


begin mathsize 14px style open square brackets C H subscript 3 C O O to the power of minus sign close square brackets equals Mol over Volume equals fraction numerator 0 comma 02 space mol over denominator left parenthesis 100 plus 100 right parenthesis cross times 10 to the power of negative sign 3 end exponent L end fraction equals 10 to the power of negative sign 1 end exponent M end style  

Lalu mencari konsentarsi ion undefined dan pH dengan rumus hidrolisis garam basa:


begin mathsize 14px style 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 open square brackets C H subscript 3 C O O to the power of minus sign close square brackets 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 5 end exponent cross times 10 to the power of negative sign 1 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 10 end exponent end root equals 10 to the power of negative sign 5 end exponent  pOH equals minus sign log open square brackets O H to the power of minus sign close square brackets pOH equals minus sign log space 10 to the power of negative sign 5 end exponent pOH equals 5  pH equals 14 minus sign pOH pH equals 14 minus sign 5 pH equals 9 end style 


Jadi, jawaban yang paling tepat adalah E.space 

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