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Pertanyaan

Konsentrasi larutan HCl adalah ...

Konsentrasi larutan HCl adalah ...

  1. 0,100 M 

  2. 0, 105 M

  3. 0, 210 M

  4. 0, 220 M

  5. 0, 400 M

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M. Firdhaus

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Pembahasan

V o l u m e space r a t a minus r a t a space H C l equals fraction numerator V o l space H C l space 1 plus V o l space H C l space 2 plus V o l space H C l space 3 over denominator B a n y a k n y a space p e r c o b a a n end fraction  equals fraction numerator 20 m L plus 20 m L plus 20 m L over denominator 3 end fraction equals 20 m L  V o l u m e space r a t a minus r a t a space B a left parenthesis O H right parenthesis subscript 2  equals space fraction numerator V o l B a left parenthesis O H right parenthesis subscript 2 1 plus V o l B a left parenthesis O H right parenthesis subscript 2 2 plus V o l B a left parenthesis O H right parenthesis subscript 2 3 over denominator B a n y a k n y a space p e r c o b a a n end fraction  equals fraction numerator 21 m L plus 22 m L plus 20 m L over denominator 3 end fraction equals 21 m L    U n t u k space m e n e n t u k a n space k o n s e n t r a s i space H C l comma space d i g u n a k a n space r u m u s space t i t r a s i space a s a m space b a s a space y a i t u colon  a space x space V a space x space M a equals b space x space V b space x space M b    D i m a n a colon  a equals J u m l a h space H space d a l a m space a s a m  V a equals V o l u m e space a s a m space left parenthesis m L right parenthesis  M a equals M o l a r i t a s divided by k o n s e n t r a s i space a s a m left parenthesis M right parenthesis  B equals J u m l a h space O H space d a l a m space b a s a space  V b equals V o l u m e space b a s a space left parenthesis m L right parenthesis  M b equals M o l a r i t a s divided by k o n s e n t r a s i space b a s a left parenthesis M right parenthesis    M a k a  a space x space V a space x space M a equals b space x space V b space x space M b  1 space x space 20 space m L space x space M a equals 2 space x space 21 m space L space x space 0 comma 1 M  M a equals fraction numerator 4 comma 2 M over denominator 20 end fraction  space space space space space space equals 0 comma 21 M

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Sebanyak 10 mL larutan HCl 0 , 1 M dititrasi dengan larutan NaOH 0 , 2 M . Maka volume NaOH yang dibutuhkan sampai tercapai titik ekivalen sebanyak ...

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