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Berapa gram HCl (Mr​ = 36,5) terlarut dalam 1 L larutan HCl dengan pH = 6.

Pertanyaan

Berapa gram HCl (M subscript r = 36,5) terlarut dalam 1 L larutan HCl dengan pH = 6.

Pembahasan Soal:

Penentuan massa HCl dapat dihitung berdasrkan data pH dan konsentrasi molarnya.

table attributes columnalign right center left columnspacing 0px end attributes row pH equals 6 row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign pH end exponent end cell row blank equals cell 10 to the power of negative sign 6 end exponent end cell row blank equals cell 0 comma 000001 end cell row blank blank blank row cell H Cl end cell rightwards arrow cell H to the power of plus sign and Cl to the power of minus sign space left parenthesis space a space colon space 1 right parenthesis end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell a cross times M end cell row M equals cell fraction numerator open square brackets H to the power of plus sign close square brackets over denominator a end fraction end cell row blank equals cell fraction numerator 0 comma 000001 over denominator 1 end fraction end cell row blank equals cell 0 comma 000001 space M end cell row blank blank blank row M equals cell m over Mr cross times 1000 over V end cell row cell 0 comma 000001 end cell equals cell fraction numerator m over denominator 36 comma 5 end fraction cross times 1000 over 1000 end cell row m equals cell fraction numerator 0 comma 000001 cross times 36 comma 5 cross times 1000 over denominator 1000 end fraction end cell row blank equals cell 0 comma 000037 space g end cell end table 

Jadi, massa HCl yang terlarut dalam 1 L larutan adalah 0,000037 gram.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. MT

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Berapa gram CH3​COOH (Mr​ = 60) terlarut dalam 5 L larutan dengan pH = 4, Ka​=10−5.

Pembahasan Soal:

Penentuan massa C H subscript 3 C O O H dapat dihitung berdasarkan data pH dan konsentrasi molarnya.

table attributes columnalign right center left columnspacing 0px end attributes row pH equals 4 row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign pH end exponent end cell row blank equals cell 10 to the power of negative sign 4 end exponent end cell row blank blank blank 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 M equals cell open square brackets H to the power of plus sign close square brackets squared over K subscript a end cell row blank equals cell left parenthesis 10 to the power of negative sign 4 end exponent right parenthesis squared over 10 to the power of negative sign 5 end exponent end cell row blank equals cell 10 to the power of negative sign 8 end exponent over 10 to the power of negative sign 5 end exponent end cell row blank equals cell 0 comma 001 space M end cell row M equals cell m over Mr cross times 1000 over V end cell row cell 0 comma 001 end cell equals cell m over 60 cross times 1000 over 5000 end cell row m equals cell fraction numerator 0 comma 001 cross times 60 cross times 5000 over denominator 1000 end fraction end cell row blank equals cell 0 comma 3 space g end cell end table  

Jadi, massa C H subscript 3 C O O H yang terlarut dalam 5 L larutan adalah 0,3 gram.

0

Roboguru

Larutan asam asetat Ka​=2×10−5 yang mempunyai pH sama dengan larutan asam klorida 1×10−3M mempunyai konsentrasi sebesar ....

Pembahasan Soal:

Ionisasi asam klorida : begin mathsize 14px style H Cl yields H to the power of plus sign and Cl to the power of minus sign end style

senyawa asam klorida memiliki jumlah ion begin mathsize 14px style H to the power of plus sign space end styleadalah 1, sehingga begin mathsize 14px style a equals 1 end style.

Menghitung pH asam klorida

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals a cross times open square brackets H Cl 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 end exponent open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 3 end exponent end style  

begin mathsize 14px style pH equals minus sign log open square brackets H to the power of plus sign close square brackets pH equals minus sign log open square brackets 10 to the power of negative sign 3 end exponent close square brackets pH equals 3 end style 

nilai begin mathsize 14px style pH space H Cl double bond C H subscript 3 C O O H equals 3 end style.

Menghitung konsentrasi asam asetat

begin mathsize 14px style 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 Ka cross times open square brackets C H subscript 3 C O O H close square brackets end root end cell row cell 10 to the power of negative sign 3 end exponent end cell equals cell square root of 2 cross times 10 to the power of negative sign 5 end exponent cross times open square brackets C H subscript 3 C O O H close square brackets end root end cell row cell open parentheses 10 to the power of negative sign 3 end exponent close parentheses squared end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times open square brackets C H subscript 3 C O O H close square brackets end cell row cell open square brackets C H subscript 3 C O O H close square brackets end cell equals cell fraction numerator 10 to the power of negative sign 6 end exponent over denominator 2 cross times 10 to the power of negative sign 5 end exponent end fraction end cell row cell open square brackets C H subscript 3 C O O H close square brackets end cell equals cell 0 comma 05 space M end cell end table end style 

Dengan demikian, jawaban yang benar adalah E.

0

Roboguru

Sebanyak 10 L larutan yang mengandung 0,1molH2​SO4​ memiliki pH sebesar ....

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell M space a s a m space end cell equals cell space fraction numerator m o l space a s a m over denominator V o l u m e end fraction space equals space fraction numerator 0 comma 1 space m o l over denominator 10 space L end fraction space space equals space 0 comma 01 space M space end cell row blank blank cell H subscript 2 S O subscript 4 space m e r u p a k a n space a s a m space k u a t comma space s e h i n g g a space colon space end cell row cell left square bracket H plus right square bracket space end cell equals cell space M subscript a s a m end subscript space x space v a l e n s i space a s a m space space space space space space space space space space space end cell row blank equals cell space 0 comma 01 space M space x space 2 space equals space 0 comma 02 space M space end cell row cell p H space end cell equals cell space minus space log space left square bracket H to the power of plus right square bracket space space space space space space space space space space end cell row blank equals cell space minus space log space 0 comma 02 space space space space space space space space space space end cell row blank equals cell space 2 space – space log space 2 end cell end table

0

Roboguru

Larutan asam klorida dalam air dengan pH = 2 akan berubah menjadi pH = 3 jika diencerkan sebanyak ...

Pembahasan Soal:

Asam klorida (HCl) merupakan asam kuat makan untuk mencari konsentrasi ion Hjika diketahui pHnya, maka menggunakan persamaan:

space space space space pH space equals space minus sign space log space open square brackets H to the power of plus sign close square brackets space space space space space space space 2 space equals space minus sign space log space open square brackets H to the power of plus sign close square brackets space space open square brackets H to the power of plus sign close square brackets space equals space 1 space x space 10 to the power of negative sign 2 end exponent space space open square brackets H to the power of plus sign close square brackets space equals space M space x space a M space x space a space equals open square brackets H to the power of plus sign close square brackets M space x space 1 space equals space 1 space x space 10 to the power of negative sign 2 end exponent space space space space space space M space equals space 1 space x space 10 to the power of negative sign 2 end exponent  Setelah space pengenceran space space space space pH space equals space minus sign space log space open square brackets H to the power of plus sign close square brackets space space space space space space space 3 space equals space minus sign space log space open square brackets H to the power of plus sign close square brackets space space open square brackets H to the power of plus sign close square brackets space equals space 1 space x space 10 to the power of negative sign 3 end exponent space space open square brackets H to the power of plus sign close square brackets space equals space M space x space a M space x space a space equals open square brackets H to the power of plus sign close square brackets M space x space 1 space equals space 1 space x space 10 to the power of negative sign 3 end exponent space space space space space space M space equals space 1 space x space 10 to the power of negative sign 3 end exponent 
 

Mencari volume pengenceran menggunakan persamaan:

space space space space space space space space space space V subscript 1 space x space M subscript 1 space equals space V subscript 2 space x space M subscript 2 V subscript 1 space x space 1 space x space 10 to the power of negative sign 2 end exponent space equals space V subscript 2 space x space 1 space x space 10 to the power of negative sign 3 end exponent space space space space space space space space space space space space space space space space space space space V subscript 2 space end subscript equals space fraction numerator V subscript 1 space x space 1 space x space 10 to the power of negative sign 2 end exponent over denominator 1 space x space 10 to the power of negative sign 3 end exponent end fraction space space space space space space space space space space space space space space space space space space space V subscript 2 space end subscript equals space 10 space V subscript 1 

Pengenceran dilakukan 10 kali volume awal.

Jadi, jawaban yang benar adalah E.   

1

Roboguru

Konsentrasi ion hidrogen dalam larutan yang pH-nya = 3 – log 2 adalah ... .

Pembahasan Soal:

pH=log[H+]3log2=log[H+][H+]=2×103M

Jadi, jawaban yang benar adalah C.

0

Roboguru

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