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Tentukan pH larutan: Larutan asam sulfat 0,03 M sebanyak 300 mL 2 mol asam nitrat dalam 4 liter air 8 gram asam bromida dilarutkan dalam 250 mL air

Pertanyaan

Tentukan pH larutan:

  1. Larutan asam sulfat 0,03 M sebanyak 300 mL
  2. 2 mol asam nitrat dalam 4 liter air
  3. 8 gram asam bromida dilarutkan dalam 250 mL air begin mathsize 14px style left parenthesis Ar space H equals 1 comma space Br equals 80 right parenthesis end stylespace 

Pembahasan Soal:

1. Larutan asam sulfat 0,03 M sebanyak 300 mL

H subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 H to the power of plus sign left parenthesis italic a italic q right parenthesis plus S O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis 

Konsentrasi ion H to the power of plus sign dapat ditentukan secara stoikometri yaitu sebagai berikut:

open square brackets H to the power of plus sign close square brackets equals a cross times M subscript a open square brackets H to the power of plus sign close square brackets equals 2 cross times 0 comma 03 open square brackets H to the power of plus sign close square brackets equals 0 comma 06 space M open square brackets H to the power of plus sign close square brackets equals 6 cross times 10 to the power of negative sign 2 end exponent space M  pH equals minus sign log space open square brackets H to the power of plus sign close square brackets pH equals minus sign log space open square brackets 6 cross times 10 to the power of negative sign 2 end exponent space M close square brackets pH equals 2 minus sign log space 6 

Jadi pH larutan asam sulfat 0,03 M sebanyak 300 mL adalah bold 2 bold minus sign bold log bold space bold 6
 

2. Larutan 2 mol asam nitrat dalam 4 liter air

H N O subscript 3 left parenthesis italic a italic q right parenthesis yields H to the power of plus sign left parenthesis italic a italic q right parenthesis plus N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis  

Konsentrasi ion H to the power of plus sign dapat ditentukan secara stoikometri yaitu sebagai berikut:

open square brackets H N O subscript 3 close square brackets equals fraction numerator mol space H N O subscript 3 over denominator V space larutan end fraction open square brackets H N O subscript 3 close square brackets equals fraction numerator 2 space mol over denominator 4 space L end fraction open square brackets H N O subscript 3 close square brackets equals 0 comma 5 space M  open square brackets H to the power of plus sign close square brackets equals 1 cross times 0 comma 5 space M open square brackets H to the power of plus sign close square brackets equals 0 comma 5 space M open square brackets H to the power of plus sign close square brackets equals 5 cross times 10 to the power of negative sign 1 end exponent space M  pH equals minus sign log space open square brackets H to the power of plus sign close square brackets pH equals minus sign log space open square brackets 5 cross times 10 to the power of negative sign 1 end exponent space M close square brackets pH equals 1 minus sign log space 5  

Jadi pH larutan asam nitrat adalah bold 1 bold minus sign bold log bold space bold 5.
 

3. 8 gram asam bromida dilarutkan dalam 250 mL air begin mathsize 14px style left parenthesis Ar space H equals 1 comma space Br equals 80 right parenthesis end style

H Br left parenthesis italic a italic q right parenthesis yields H to the power of plus sign left parenthesis italic a italic q right parenthesis plus Br to the power of minus sign left parenthesis italic a italic q right parenthesis   

Konsentrasi ion H to the power of plus sign dapat ditentukan secara stoikometri yaitu sebagai berikut:

open square brackets H Br close square brackets equals fraction numerator massa space H Br over denominator M subscript r space H Br end fraction cross times fraction numerator 1000 over denominator V space larutan end fraction open square brackets H Br close square brackets equals fraction numerator 8 space gram over denominator 81 space gram space mol to the power of negative sign 1 end exponent end fraction cross times fraction numerator 1000 over denominator 250 space mL end fraction open square brackets H Br close square brackets equals 3 comma 95 cross times 10 to the power of negative sign 1 end exponent space M  open square brackets H to the power of plus sign close square brackets equals 1 cross times 3 comma 95 cross times 10 to the power of negative sign 1 end exponent space M open square brackets H to the power of plus sign close square brackets equals 3 comma 95 cross times 10 to the power of negative sign 1 end exponent space M  pH equals minus sign log space open square brackets H to the power of plus sign close square brackets pH equals minus sign log space open square brackets 3 comma 95 cross times 10 to the power of negative sign 1 end exponent close square brackets pH equals 1 minus sign log space 3 comma 95    

Jadi pH larutan asam bromida adalah bold 1 bold minus sign bold log bold space bold 3 bold comma bold 95.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Juni 2021

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