Roboguru

Penambahan 20 mL larutan natrium fosfat 0,2 M ke dalam 100 mL  mengakibatkan ...

Pertanyaan

Penambahan 20 mL larutan natrium fosfat 0,2 M ke dalam 100 mL begin mathsize 14px style H subscript 3 P O subscript 4 space 0 comma 1 space M space left parenthesis K subscript a equals 7 cross times 10 to the power of negative sign 3 end exponent right parenthesis end style mengakibatkan ...

  1. Kenaikan pH sebesar 5,05

  2. Kenaikan pH sebesar 3,03

  3. Penurunan pH sebesar 3,05

  4. Penurunan pH sebesar 5,05undefined

  5. Penurunan pH sebesar 7,05

Pembahasan Soal:

Pertama kita tentukan dulu pH larutan begin mathsize 14px style H subscript 3 P O subscript 4 space 0 comma 1 space M space left parenthesis K subscript a equals 7 cross times 10 to the power of negative sign 3 end exponent right parenthesis end style sebelum penambahan:


begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K subscript a cross times M subscript H subscript 3 P O subscript 4 end subscript end root open square brackets H to the power of plus sign close square brackets equals square root of 7 cross times 10 to the power of negative sign 3 end exponent cross times 10 to the power of negative sign 1 end exponent end root open square brackets H to the power of plus sign close square brackets equals square root of 7 cross times 10 to the power of negative sign 4 end exponent end root open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 2 end exponent cross times square root of 7  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 2 end exponent cross times square root of 7 right parenthesis pH equals 2 minus sign log space square root of 7 pH almost equal to 1 comma 58 end style   


Setelah penambahan 20 mL larutan garam natrium fosfat 0,2 M akan membentuk larutan penyangga dengan pH:


begin mathsize 14px style open square brackets H to the power of plus sign close square brackets double bond K subscript a cross times n subscript H subscript 3 P O subscript 4 end subscript over n subscript Na subscript 3 P O subscript 4 end subscript open square brackets H to the power of plus sign close square brackets equals italic 7 cross times 10 to the power of negative sign 3 end exponent cross times fraction numerator 0 comma 1 space M cross times 100 space mL over denominator 0 comma 2 space M cross times 20 space mL end fraction open square brackets H to the power of plus sign close square brackets equals 1 comma 75 cross times 10 to the power of negative sign 2 end exponent  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 1 comma 75 cross times 10 to the power of negative sign 2 end exponent right parenthesis pH equals 2 minus sign log space 1 comma 75 pH almost equal to 1 comma 76 end style 


Terjadi kenaikan nilai pH sebesar 0,18 menjadi 1,76 setelah penambahan 20 mL larutan natrium fosfat 0,2 M.


Jadi, tidak ada jawaban yang tepat.space space space space

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 April 2021

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved