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Sebanyak 100 ml larutan  0,1 M dicampur dengan 100...

Sebanyak 100 ml larutan begin mathsize 14px style H subscript 2 S O subscript 4 end style 0,1 M dicampur dengan 100 ml larutan begin mathsize 14px style N H subscript 3 end style 0,2 M. Jika begin mathsize 14px style K subscript b space N H subscript 3 equals 1 cross times 10 to the power of negative sign 5 end exponent end style maka pH campuran yang terbentuk adalah...space 

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 H subscript 2 S O subscript 4 double bond M cross times V Mol space H subscript 2 S O subscript 4 equals 0 comma 1 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 H subscript 2 S O subscript 4 equals 0 comma 01 space mol  Mol space N H subscript 3 double bond M cross times V Mol space N H subscript 3 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 N H subscript 3 equals 0 comma 02 space mol end style 


Reaksi asam basa:
 

begin mathsize 14px style N H subscript 3 and H subscript 2 O yields N H subscript 4 O H 0 comma 02 space mol space space space space space space space space 0 comma 02 space mol end style  

 

space space space space space space space H subscript 2 S O subscript 4 space space space space plus space space space space 2 N H subscript 4 O H space space space rightwards arrow space space space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 space space plus space space 2 H subscript 2 O m space space space 0 comma 01 space mol space space space space space space space space space 0 comma 02 space mol space space space space space space space space space space space space space space space space space space minus sign r space space space space space 0 comma 01 space mol space space space space space space space space space 0 comma 02 space mol space space space space space space space space space space space space space space 0 comma 01 space mol s space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space space space space space space 0 comma 01 space mol space

Jadi setelah reaksi, terbentuk larutan garam begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end style sebanyak 0,01 mol. Larutan garam ini akan terionisasi membentuk begin mathsize 14px style N H subscript 4 to the power of plus end style dan begin mathsize 14px style S O subscript 4 to the power of 2 minus sign end exponent end style. undefined akan terhidrolisis oleh air sehingga terbentuk ion begin mathsize 14px style H to the power of plus sign end style.
 

begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript yields 2 N H subscript 4 to the power of plus plus S O subscript 4 to the power of 2 minus sign end exponent 0 comma 01 space mol space space space space space space 0 comma 02 space mol N H subscript 4 to the power of plus plus H subscript 2 O equilibrium N H subscript 4 O H and H to the power of plus sign 0 comma 02 space mol end style 


Kemudian mengubah mol menjadi konsentrasi,
 

begin mathsize 14px style left square bracket N H subscript 4 to the power of plus right square bracket 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 konsentrasi ion undefined dan pH dengan rumus hidrolisis garam asam:
 

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K subscript italic w over K subscript italic b cross times left square bracket N H subscript 4 to the power of plus right square bracket end root open square brackets H to the power of plus 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 left square bracket 10 to the power of negative sign 1 end exponent right square bracket end root open square brackets H to the power of plus 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 pH space space equals minus sign space log open square brackets H to the power of plus sign close square brackets pH space space equals minus sign space log space 10 to the power of negative sign 5 end exponent pH space space equals 5 end style 


Jadi, pH campuran yang dihasilkan adalah 5.space  

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