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Jika 100 mL larutan asam klorida dengan pH = 2 dicampurkan pada 100 mL larutan natrium hidroksida dengan pH = 10, akan diperoleh larutan dengan ....

Pertanyaan

Jika 100 mL larutan asam klorida dengan pH = 2 dicampurkan pada 100 mL larutan natrium hidroksida dengan pH = 10, akan diperoleh larutan dengan .... 

  1. pH = 3 

  2. pH = 6 

  3. 2 < pH < 6 

  4. 3 < pH < 6 

  5. 6 < pH < 1 

Pembahasan Soal:

Menentukan mol dari masing-masing zat pereaksi:

H Cl space space space space space space space space space space space space space space pH space equals space 2 space minus sign space log space open square brackets H to the power of plus sign close square brackets space equals 2 space space space space space space space space space space space open square brackets H to the power of plus sign close square brackets space equals space 10 to the power of negative sign 2 end exponent space space space space space space space space space space M space x space a space equals space 10 to the power of negative sign 2 end exponent space space space space space space space space space space M space x space 1 space equals space 10 to the power of negative sign 2 end exponent space space space space space space space space space space space space space space space space M space equals space 10 to the power of negative sign 2 end exponent  mol space equals space M space x space V mol space equals space 10 to the power of negative sign 2 end exponent space x space 100 space mL mol space equals space 1 space mmol  Na O H pH space equals space 10 pOH space equals space 14 space minus sign 10 pOH space equals space 4 space space space space space space space space space space space space space space pOH space equals space 4 space minus sign space log space open square brackets O H to the power of minus sign close square brackets space equals 4 space space space space space space space space space space space open square brackets O H to the power of minus sign close square brackets space equals space 10 to the power of negative sign 4 end exponent space space space space space space space space space space space M space x space a space equals space 10 to the power of negative sign 4 end exponent space space space space space space space space space space space M space x space 1 space equals space 10 to the power of negative sign 4 end exponent space space space space space space space space space space space space space space space space M space equals space 10 to the power of negative sign 4 end exponent  mol space equals space M space x space V mol space equals space 10 to the power of negative sign 4 end exponent space x space 100 space mL mol space equals space 10 to the power of negative sign 2 end exponent space mmol 

Pada larutan terdapat 0,99 mmol HCl, maka larutan bersifat asam, untuk menghitung pH larutan maka menggunakan persamaan:

M space equals space n over V subscript total M space equals space fraction numerator 0 comma 99 space mmol over denominator 200 space mL end fraction M space equals 4 comma 95 space x space 10 to the power of negative sign 3 end exponent  open square brackets H to the power of plus sign close square brackets space equals space M space x space a open square brackets H to the power of plus sign close square brackets space equals space 4 comma 95 space x space 10 to the power of negative sign 3 end exponent space x space 1 open square brackets H to the power of plus sign close square brackets space equals space 4 comma 95 space x space 10 to the power of negative sign 3 end exponent  pH space equals space minus sign space log space open square brackets H to the power of plus sign close square brackets space pH space equals space minus sign space log space 4 comma 95 space x space 10 to the power of negative sign 3 end exponent pH space equals 3 space minus sign space log space 4 comma 95 pH space equals 2 comma 3 

Jadi, jawaban yang benar adalah C. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Kartika

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Hitung pH larutan campuran berikut :

Pembahasan Soal:

Untuk menentukan pH campuran asam kuat dengan asam lemah tidak dapat dilakukan dengan cara meghitung begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end stylesetiap asam dan menjumlahkan keduanya kemudian menghitung pH dari undefined campuran itu. Perlu dipertimbangkan konsep kesetimbangan larutan asam lemah. Oleh sebab itu, rumus yang digunakan untuk mencari undefinedcampuran asam kuat dan asam lemah adalah sebagai berikut:

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals fraction numerator M subscript 1 plus-or-minus square root of open parentheses M subscript 1 close parentheses squared plus 4 KaM subscript 2 end root over denominator 2 end fraction end style 

begin mathsize 14px style M subscript 1 end style = konsentrasi asam kuat, yaitu begin mathsize 14px style H N O subscript 3 end style sebesar 0,04 M
begin mathsize 14px style M subscript 2 end style = konsentrasi asam lemah, yaitu begin mathsize 14px style C H subscript 3 C O O H end style sebesar 0,1 M
Ka = tetapan kesetimbangan asam lemah, yaitu sebesar begin mathsize 14px style 1 cross times 10 to the power of negative sign 5 end exponent end style 

Perhitungan nilai undefined campuran adalah sebagai berikut:

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 fraction numerator 0 comma 04 plus-or-minus square root of left parenthesis 0 comma 04 right parenthesis squared plus 4 open parentheses 1 cross times 10 to the power of negative sign 5 end exponent close parentheses open parentheses 0 comma 1 close parentheses end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 0 comma 04 plus-or-minus square root of 0 comma 0016 plus open parentheses 4 cross times 10 to the power of negative sign 6 end exponent close parentheses end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 0 comma 04 plus-or-minus square root of 1 comma 604 cross times 10 to the power of negative sign 3 end exponent end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 0 comma 04 plus-or-minus 0 comma 04005 over denominator 2 end fraction end cell end table end style 
Karena undefined tidak mungkin bernilai negatif maka:
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 fraction numerator 0 comma 04 plus 0 comma 04005 over denominator 2 end fraction end cell row blank equals cell 0 comma 040025 end cell end table end style 
Sehingga nilai pH campuran adalah sebagai berikut:
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log open parentheses 0 comma 040025 close parentheses end cell row blank equals cell 1 comma 3977 end cell end table end style 

Dengan demikian, pH campuran senyawa tersebut adalah 1,3977.

0

Roboguru

Hitung pH larutan campuran berikut :

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell H Cl end cell equals cell 50 space mL space x space 0 comma 2 space M equals 10 space mmol end cell row cell Na O H end cell equals cell 50 space mL space x space 0 comma 1 space M equals 5 space mmol end cell end table end style 

 

Jadi, setelah reaksi masih tersisa 5 mmol dalam 100 mL larutan, maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell H Cl end cell equals cell fraction numerator 5 space mmol over denominator 100 space mL end fraction end cell row blank equals cell 0 comma 05 space M end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 0 comma 05 space M end cell row pH equals cell negative sign log open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log left parenthesis 0 comma 05 right parenthesis end cell row blank equals cell 2 minus sign log space 5 end cell row blank equals cell 1 comma 3 end cell end table end style

Jadi, pH yang dihasilkan dari campuran reaksi di atas adalah 1,3.

0

Roboguru

Hitung pH larutan campuran berikut :

Pembahasan Soal:

Untuk mencari pH campuran, digunakan rumus berikut:

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets subscript 1 cross times V subscript 1 end subscript plus open square brackets H to the power of plus sign close square brackets subscript 2 cross times V subscript 2 equals open square brackets H to the power of plus sign close square brackets subscript camp cross times V subscript camp end style 

Maka dicari terlebih dahulu konsentrasi begin mathsize 14px style H to the power of plus sign end style masing-masing.

Larutan begin mathsize 14px style H Cl end style 

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 a cross times M subscript a end cell row blank equals cell 1 cross times 0 comma 1 end cell row blank equals cell 10 to the power of negative sign 1 end exponent end cell end table end style 

Larutan begin mathsize 14px style H subscript 2 S O subscript 4 end style 

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 a cross times M subscript a end cell row blank equals cell 2 cross times 0 comma 2 end cell row blank equals cell 4 cross times 10 to the power of negative sign 1 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis open square brackets H to the power of plus sign close square brackets subscript 1 cross times V subscript 1 right parenthesis plus left parenthesis open square brackets H to the power of plus sign close square brackets subscript 2 cross times V subscript 2 right parenthesis end cell equals cell open square brackets H to the power of plus sign close square brackets subscript camp cross times V subscript camp end cell row cell open square brackets H to the power of plus sign close square brackets subscript camp end cell equals cell fraction numerator left parenthesis open square brackets H to the power of plus sign close square brackets subscript 1 cross times V subscript 1 right parenthesis plus left parenthesis open square brackets H to the power of plus sign close square brackets subscript 2 cross times V subscript 2 right parenthesis over denominator V subscript camp end fraction end cell row blank equals cell fraction numerator left parenthesis 10 to the power of negative sign 1 end exponent cross times 100 right parenthesis space plus space left parenthesis 4 cross times 10 to the power of negative sign 1 end exponent cross times 100 right parenthesis over denominator 200 end fraction end cell row blank equals cell fraction numerator 10 plus 40 over denominator 200 end fraction end cell row blank equals cell 0 comma 25 end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 2 comma 5 cross times 10 to the power of negative sign 1 end exponent end cell row blank equals cell 1 minus sign log space 2 comma 5 end cell end table end style 

Jadi, pH campuran asam tersebut adalah 1 - log 2,5. 

0

Roboguru

Jika ke dalam 1 liter asam lemah HZ 0,3 M (Ka = 10-5) dilarutkan 8 gram NaOH padat (Mr = 40), maka pH campuran adalah....

Pembahasan Soal:

Reaksi yang terjadi ialah:
 

 
 

Hasil reaksi sisa asam lemah dan garam dengan masing0masing berjumlah 0,1 mol dan 0,2 mol. Oleh karena itu terbentuk sistem penyangga asam lemah.
 

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 Ka cross times fraction numerator n space asam space lemah over denominator n space garam end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction end cell row blank equals cell 5 cross times 10 to the power of negative sign 6 end exponent end cell row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 5 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 6 space minus sign space log space 5 end cell end table 
 

Jadi, dapat disimpulkan jawaban yang tepat adalah D.space

0

Roboguru

Sebanyak 200 mL larutan penyangga mengandung  dan  masing-masing 0,05 M. Tentukan pH larutan setelah 2 mL NaOH 0,05 M! ()

Pembahasan Soal:

Pada saat penambahan NaOH, komponen yang bereaksi adalah komponen asam konjugasinya, yaitu begin mathsize 14px style N H subscript 4 Cl end style.


begin mathsize 14px style Mol space N H subscript 3 double bond M cross times V equals 200 mL cross times 0 comma 05 M equals 10 space mmol Mol space N H subscript 4 Cl double bond M cross times V equals 200 mL cross times 0 comma 05 M equals 10 space mmol Mol space Na O H double bond M cross times V equals 2 mL cross times 0 comma 05 M equals 0 comma 1 space mmol end style 

 
 

Setelah penambahan NaOH, mol yang bersisa masih terdiri atas basa lemah (begin mathsize 14px style N H subscript 3 end style) dan garamnya (begin mathsize 14px style N H subscript 4 Cl end style), sehingga larutan tersebut bersifat larutan penyangga basa.
 

begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets double bond K subscript b cross times fraction numerator mol space basa over denominator mol space garam end fraction equals 10 to the power of negative sign 5 end exponent cross times fraction numerator 10 comma 1 over denominator 9 comma 9 end fraction equals 1 comma 02 cross times 10 to the power of negative sign 5 end exponent pOH equals minus sign log space open square brackets O H to the power of minus sign close square brackets equals minus sign log space 1 comma 02 cross times 10 to the power of negative sign 5 end exponent equals 5 pH equals 14 minus sign pOH equals 14 minus sign 5 equals 9 end style 
 

Jadi, pH campuran larutan tersebut adalah 9.space 

0

Roboguru

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