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Larutan 50 ml HCI 0,2 M ditambah 50 ml larutan 0,2 M () sesuai dengan reaksi: pH larutan yang terjadi adalah....

Larutan 50 ml HCI 0,2 M ditambah 50 ml larutan begin mathsize 14px style N H subscript 3 end style 0,2 M (begin mathsize 14px style Kb equals 1 cross times 10 to the power of negative sign 5 end exponent end style) sesuai dengan reaksi:
 

begin mathsize 14px style H Cl left parenthesis italic a italic q right parenthesis plus N H subscript 3 left parenthesis italic a italic q right parenthesis yields N H subscript 4 Cl left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses end style 

 
pH larutan yang terjadi adalah....undefined 

  1. 5undefined 

  2. 7undefined 

  3. 9undefined 

  4. 10undefined 

  5. 12undefined 

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 Cl double bond M cross times V Mol space H Cl equals 0 comma 2 begin inline style bevelled mol over L end style cross times 50 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 Cl 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 50 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 01 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 01 space mol space space space space space space space 0 comma 01 space mol end style  

 

Jadi setelah reaksi, terbentuk larutan garam begin mathsize 14px style N H subscript 4 Cl 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 Cl to the power of minus sign end styleundefined akan menghidrolisis air sehingga terbentuk ion begin mathsize 14px style H to the power of plus sign end style.


begin mathsize 14px style N H subscript 4 Cl yields N H subscript 4 to the power of plus plus Cl to the power of minus sign 0 comma 01 space mol space 0 comma 01 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 01 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 01 space mol over denominator left parenthesis 50 plus 50 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 asam:


begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K subscript w over K subscript 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 10 to the power of negative sign 1 end exponent 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 equals minus sign log open square brackets H to the power of plus sign close square brackets pH equals minus sign log space 10 to the power of negative sign 5 end exponent pH equals 5 end style 


Jadi, jawaban yang paling tepat adalah A.space 

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