pH larutan BaF2​ 0,02 M (Ka​=10−4) adalah ....

Pertanyaan

pH larutan begin mathsize 14px style Ba F subscript 2 space 0 comma 02 space M space left parenthesis K subscript a equals 10 to the power of negative sign 4 end exponent right parenthesis end style adalah ....

  1. begin mathsize 14px style 3 minus sign log space 2 end style 

  2. begin mathsize 14px style 6 minus sign log space 2 end style 

  3. begin mathsize 14px style 6 minus sign begin inline style 1 half end style log space 2 end style 

  4. begin mathsize 14px style 8 plus begin inline style 1 half end style log space 2 end style 

  5. begin mathsize 14px style 8 plus log space 2 end style

G. Suprobo

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.

Pembahasan

Larutan begin mathsize 14px style Ba F subscript 2 end style terbentuk dari reaksi asam lemah (HF) dan basa kuat begin mathsize 14px style left parenthesis Ba open parentheses O H close parentheses subscript 2 right parenthesis end style yang akan mengalami hidrolisis anion yang berasal dari asam lemah sehingga larutannya bersifat basa. 

begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets equals square root of Kw over Ka end root x space M Ka double bond tetapan space asam space lemah Kw double bond tetapan space ionisasi space air space left parenthesis 1 x 10 to the power of negative sign 14 end exponent right parenthesis M equals space molaritas space anion space garam space yang space terhidrolisis end style

Larutan begin mathsize 14px style Ba F subscript 2 end style di dalam air akan terionisasi menurut persamaan reaksi sebagai berikut :

begin mathsize 14px style Ba F subscript 2 open parentheses aq close parentheses yields Ba to the power of 2 plus sign open parentheses aq close parentheses and 2 F to the power of minus sign open parentheses aq close parentheses 0 comma 02 space M space space space space space space 0 comma 02 space M space space space space space space space 0 comma 04 space M end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Molaritas space anion space open parentheses F to the power of minus sign close parentheses end cell equals cell jumlah space ion space x space Molaritas end cell row blank equals cell 2 x 0 comma 02 space M end cell row blank equals cell 0 comma 04 space M end cell end table 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 end cell equals cell square root of fraction numerator 1 x 10 to the power of negative sign 14 end exponent over denominator 10 to the power of negative sign 4 end exponent end fraction end root x space left parenthesis 4 x 10 to the power of negative sign 2 end exponent right parenthesis end cell row blank equals cell square root of 4 x 10 to the power of negative sign 14 minus sign 2 plus 4 end exponent end root end cell row blank equals cell square root of 4 x 10 to the power of negative sign 12 end exponent end root end cell row blank equals cell 2 x 10 to the power of negative sign 6 end exponent end cell 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 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 x 10 to the power of negative sign 6 end exponent right parenthesis end cell row blank equals cell 6 minus sign log space 2 end cell row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign left parenthesis 6 minus sign log space 2 right parenthesis end cell row blank blank cell 8 plus log space 2 end cell end table end style 

Dengan demikian, pH larutan begin mathsize 14px style Ba F subscript 2 space 0 comma 02 space M space left parenthesis K subscript a equals 10 to the power of negative sign 4 end exponent right parenthesis end style adalah 8+log 2.

Jadi, jawaban yang benar adalah E.

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