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Suatu asam lemah 0,01 M memiliki pH 3,5. Konstanta asam tersebut adalah....

Suatu asam lemah begin mathsize 14px style H C O O H end style 0,01 M memiliki pH 3,5. Konstanta asam begin mathsize 14px style open parentheses K subscript a close parentheses end style tersebut adalah....

  1. 5 cross times 10 to the power of negative sign 8 end exponent 

  2. begin mathsize 14px style 1 cross times 10 to the power of negative sign 7 end exponent end style 

  3. begin mathsize 14px style 1 cross times 10 to the power of negative sign 5 end exponent end style 

  4. begin mathsize 14px style 2 cross times 10 to the power of negative sign 3 end exponent end style 

  5. begin mathsize 14px style 1 cross times 10 to the power of negative sign 2 end exponent end style 

Jawaban:

Diketahui:
Suatu asam lemah begin mathsize 14px style H C O O H end style 0,01 M
pH = 3,5

Ditanya:
Konstanta asam begin mathsize 14px style open parentheses K subscript a close parentheses end style

Penyelesaian:
Untuk menentukan konstanta asam begin mathsize 14px style open parentheses K subscript a close parentheses end style begin mathsize 14px style H C O O H end style, digunakan rumus:

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K subscript a cross times M end root end style

Adapun langkah-langkahnya yaitu:

1. Menghitung konsentrasi H+ dari pH yang telah diketahui

pH equals 3 comma 5 pH equals minus sign log space open square brackets H to the power of plus sign close square brackets 3 comma 5 equals minus sign log space open square brackets H to the power of plus sign close square brackets open square brackets H to the power of plus sign close square brackets equals 1 cross times 10 to the power of negative sign 3 comma 5 end exponent space M

2. Menghitung Ka asam lemah

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 K subscript a cross times M end root end cell row cell 1 cross times 10 to the power of negative sign 3 comma 5 end exponent end cell equals cell square root of K subscript a cross times left parenthesis 1 cross times 10 to the power of negative sign 2 end exponent right parenthesis end root end cell row cell left parenthesis 1 cross times 10 to the power of negative sign 3 comma 5 end exponent right parenthesis squared end cell equals cell K subscript a end subscript cross times left parenthesis 1 cross times 10 to the power of negative sign 2 end exponent right parenthesis end cell row cell 1 cross times 10 to the power of negative sign 7 end exponent end cell equals cell K subscript a cross times left parenthesis 1 cross times 10 to the power of negative sign 2 end exponent right parenthesis end cell end table 

K subscript italic a equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 1 cross times 10 to the power of negative sign 7 end exponent over denominator 1 cross times 10 to the power of negative sign 2 end exponent end fraction end cell end table K subscript a equals 1 cross times 10 to the power of negative sign 5 end exponent    

Jadi, jawaban yang tepat adalah C.

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