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Campuran yang terdiri dari 100 mL HF 0,1 M (Ka​=2×10−5) dan 100 ml NaF 0,2 M akan memiliki pH sebesar ...

Pertanyaan

Campuran yang terdiri dari 100 mL H F 0,1 M left parenthesis K subscript a equals 2 cross times 10 to the power of negative sign 5 end exponent right parenthesis dan 100 ml Na F 0,2 M akan memiliki pH sebesar ...space 

  1. 5space  

  2. 5 plus log space 9space  

  3. 9 minus sign log space 5space  

  4. 9space  

  5. 9 plus log space 5space  

Pembahasan Soal:

Larutan Penyangga Asam

Campuran asam lemah H F dan basa konjugasinya Na F akan membentuk larutan penyangga dengan reaksi sebagai berikut:

H F left parenthesis italic a italic q right parenthesis equilibrium F to the power of minus sign left parenthesis italic a italic q right parenthesis plus H to the power of plus sign left parenthesis italic a italic q right parenthesis Na F left parenthesis italic a italic q right parenthesis yields F to the power of minus sign left parenthesis italic a italic q right parenthesis plus Na to the power of plus sign left parenthesis italic a italic q right parenthesis 

Campuran tersebut merupakan larutan penyangga asam karena menghasilkan ion H to the power of plus sign, oleh karena itu konsentrasi H to the power of plus sign dan pH dapat ditentukan sebagai berikut:

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 K subscript a cross times fraction numerator mol space asam space lemah over denominator mol space basa space konjugasi end fraction end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell K subscript a cross times fraction numerator mol space H F over denominator mol space Na F end fraction end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell K subscript a cross times fraction numerator V space H F cross times M space H F over denominator V space Na F cross times M space Na F end fraction end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 100 space mL cross times 0 comma 1 space M over denominator 100 space mL cross times 0 comma 2 space M end fraction end cell 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 5 end exponent space M end cell row blank blank blank row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row pH equals cell negative sign log space open square brackets 10 to the power of negative sign 5 end exponent space M close square brackets end cell row pH equals 5 end table

Jadi pH campuran tersebut adalah 5.

Oleh karena itu, jawaban yang benar adalah A.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Rahayu

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Ke dalam 100 mL 0,1 M larutan asam asetat (Ka​=1×10−5) ditambahkan sejumlah garam natrium asetat (Mr​=82) hingga pH larutan naik menjadi 5. Massa natrium asetat yang ditambahkan adalah ...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell pH space campuran end cell equals 5 row cell H to the power of plus sign end cell equals cell 10 to the power of negative sign 5 end exponent space M 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 K subscript a cross times fraction numerator mol space C H subscript 3 C O O H over denominator mol space C H subscript 3 C O O Na end fraction end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell K subscript a cross times fraction numerator V space C H subscript 3 C O O H cross times M space C H subscript 3 C O O H over denominator begin display style fraction numerator g space C H subscript 3 C O O Na over denominator M subscript r space C H subscript 3 C O O Na end fraction end style end fraction end cell row cell 10 to the power of negative sign 5 space end exponent M end cell equals cell 10 to the power of negative sign 5 end exponent space fraction numerator begin display style 100 space mL cross times 0 comma 1 space M end style over denominator begin display style fraction numerator g space C H subscript 3 C O O Na over denominator 82 space g space mol to the power of negative sign 1 end exponent end fraction end style end fraction end cell row 1 equals cell 10 space mmol cross times fraction numerator 82 space g space mol to the power of negative sign 1 end exponent over denominator g space C H subscript 3 C O O Na end fraction end cell row cell g space C H subscript 3 C O O Na end cell equals cell 10 cross times 10 to the power of negative sign 3 end exponent space mol cross times 82 space g space mol to the power of negative sign 1 end exponent end cell row cell g space C H subscript 3 C O O Na end cell equals cell 0 comma 82 space gram end cell end table 

Jadi massa natrium asetat yang ditambahkan adalah 8,2 gram.

Oleh karena itu, jawaban yang benar adalah B.space 

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Roboguru

Magnesium nitrit yang mengandung 0,32 g atom oksigen direaksikan dengan 45 mL asam nitrit 0,2 M. Jika ke dalam sistem tersebut ditambahkan 1 mL HCl 1 M. Berapakah pH sebelum dan sesudah penambahan asa...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell m space O end cell equals cell fraction numerator indeks space O cross times A subscript r space O over denominator M subscript r space Mg open parentheses N O subscript 2 close parentheses subscript 2 end fraction cross times m space Mg open parentheses N O subscript 2 close parentheses subscript 2 end cell row cell 0 comma 32 space g end cell equals cell fraction numerator 4 cross times 16 over denominator 116 end fraction cross times m space Mg open parentheses N O subscript 2 close parentheses subscript 2 end cell row cell m space Mg open parentheses N O subscript 2 close parentheses subscript 2 end cell equals cell 0 comma 58 space g end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell n space Mg open parentheses N O subscript 2 close parentheses subscript 2 end cell equals cell m over M subscript r end cell row blank equals cell fraction numerator 0 comma 58 over denominator 116 end fraction end cell row blank equals cell 0 comma 005 space mol end cell row blank equals cell 5 space mmol end cell end table 


table attributes columnalign right center left columnspacing 0px end attributes row cell n space H N O subscript 2 end cell equals cell V cross times M end cell row blank equals cell 45 space mL cross times 0 comma 2 space M end cell row blank equals cell 9 space mmol end cell end table 


pH sebelum penambahan asam:
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 K subscript a cross times fraction numerator open square brackets H N O subscript 2 close square brackets over denominator 2 cross times left square bracket Mg open parentheses N O subscript 2 close parentheses subscript 2 right square bracket end fraction end cell row blank equals cell 4 comma 5 cross times 10 to the power of negative sign 4 end exponent cross times fraction numerator 9 space mmol over denominator 2 cross times 5 space mmol end fraction end cell row blank equals cell 4 comma 05 cross times 10 to the power of negative sign 4 end exponent end cell end table 

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 left parenthesis 4 comma 05 cross times 10 to the power of negative sign 4 end exponent right parenthesis end cell row blank equals cell 4 minus sign log space 4 comma 05 end cell end table 


pH setelah penambahan asam:

table attributes columnalign right center left columnspacing 0px end attributes row cell n space H Cl end cell equals cell V cross times M end cell row blank equals cell 1 space mL cross times 1 space M end cell row blank equals cell 1 space mmol end cell end table 

Asam yang ditambahkan akan dinetralkan oleh basa konjugasinya, yaitu magnesium nitrit.

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 K subscript a cross times fraction numerator open square brackets H N O subscript 2 close square brackets over denominator 2 cross times left square bracket Mg open parentheses N O subscript 2 close parentheses subscript 2 right square bracket end fraction end cell row blank equals cell 4 comma 5 cross times 10 to the power of negative sign 4 end exponent cross times fraction numerator 10 space mmol over denominator 2 cross times 4 comma 5 space mmol end fraction end cell row blank equals cell 5 cross times 10 to the power of negative sign 4 end exponent end cell end table 

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 left parenthesis 5 cross times 10 to the power of negative sign 4 end exponent right parenthesis end cell row blank equals cell 4 minus sign log space 5 end cell end table 


Jadi, pH sebelum penambahan asam sebesar 4 - log 4,05 dan pH setelah penambahan asam sebesar 4 - log 5.

0

Roboguru

Berapakah pH larutan, jika ke dalam 0,4 L larutan 0,1 M CH3COOH dilarutkan 0,04 mol CH3COONa? Harga Ka​CH3​COOH=10−7.

Pembahasan Soal:

Larutan asam lemah C H subscript 3 C O O H dicampurkan dengan garam C H subscript 3 C O O Na akan membentuk larutan buffer asam. pH larutan buffer tersebut dapat dihitung dengan cara:

open square brackets H to the power of plus sign close square brackets double bond K subscript a cross times fraction numerator n space C H subscript 3 C O O H over denominator n space C H subscript 3 C O O Na end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 7 end exponent cross times fraction numerator 0 comma 4 space L cross times 0 comma 1 space M over denominator 0 comma 04 space mol end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 7 end exponent cross times fraction numerator 0 comma 04 space mol over denominator 0 comma 04 space mol end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 7 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 left parenthesis 10 to the power of negative sign 7 end exponent right parenthesis pH equals 7  


Jadi, pH larutan buffer asam tersebut adalah 7.space space space space

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Roboguru

Berapa pH larutan penyangga yang terbuat dari campuran natrium asetat 0,05 M dan asam asetat 0,01 M? (Ka​CH3​COOH=1,6×10−5).

Pembahasan Soal:

Larutan penyangga yang terbentuk antara asam asetat (asam lemah) dan natrium asetat (garam) merupakan larutan penyangga asam.

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 K subscript a cross times fraction numerator left square bracket asam space asetat right square bracket over denominator left square bracket natrium space asetat right square bracket end fraction end cell row blank equals cell 1 comma 6 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 0 comma 01 over denominator 0 comma 05 end fraction end cell row blank equals cell 3 comma 2 cross times 10 to the power of negative sign 6 end exponent end cell end table 

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 left parenthesis 3 comma 2 cross times 10 to the power of negative sign 6 end exponent right parenthesis end cell row blank equals cell 6 minus sign log space 3 comma 2 end cell end table 


Jadi, pH larutan penyangga yang terbentuk sebesar 6 - log 3,2.

0

Roboguru

Sebanyak x gram HCOOH(Ka​=1,8×10−4) dan y gram  HCOOK dilarutkan dalam 1000 mL air sampai diperoleh pH=5−log9. Perbandıngan massa antara HCOOH dan  tersebut adalah ... (Mr​HCOOH=46grammol−1danMr​HCOOK...

Pembahasan Soal:

Jenis larutan penyangga tersebut adalah larutan penyangga asam karena terdiri dari asam lemah dan garamnya. Untuk mencari perbandingan masa asam dan massa garamnya maka perlu dicari dengan beberapa langkah berikut:

Mencari konsentrasi ion begin mathsize 14px style H to the power of bold plus sign end style 

begin mathsize 14px style pH equals 5 minus sign log space 9 H to the power of plus sign equals 9 cross times 10 to the power of negative sign 5 end exponent space M end style 

Mencari perbandingan massa asam dan garamnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell H to the power of plus sign end cell equals cell K subscript a cross times fraction numerator mol space asam space lemah over denominator mol space garam end fraction end cell row blank blank blank row cell H to the power of plus space end exponent end cell equals cell K subscript a cross times fraction numerator begin display style fraction numerator italic x over denominator M subscript r space H C O O H end fraction end style over denominator begin display style fraction numerator italic y over denominator M subscript r space H C O O K end fraction end style end fraction end cell row cell H to the power of plus sign end cell equals cell K subscript a cross times fraction numerator italic x over denominator M subscript r space H C O O H end fraction cross times fraction numerator M subscript r space H C O O K over denominator italic y end fraction end cell row cell 9 cross times 10 to the power of negative sign 5 space end exponent M end cell equals cell 1 comma 8 cross times 10 to the power of negative sign 4 end exponent cross times fraction numerator begin display style italic x end style over denominator begin display style 46 space g space mol to the power of negative sign 1 end exponent end style end fraction cross times fraction numerator 84 space g space mol to the power of negative sign 1 end exponent over denominator italic y end fraction end cell row 1 equals cell fraction numerator 168 x over denominator 46 y end fraction end cell row cell 46 y end cell equals cell 168 x end cell row cell italic x over italic y end cell equals cell 46 over 168 end cell row cell italic x over italic y end cell equals cell 23 over 84 end cell end table end style 

Jadi perbandingan massa begin mathsize 14px style H C O O H bold space bold colon bold space H C O O K end style adalah begin mathsize 14px style bold 23 bold space bold colon bold space bold 84 end style

Oleh karena itu, jawaban yang benar adalah C.space 

0

Roboguru

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