Roboguru

60mL larutan  0,2M yang mempunyai pH=3-log 2 dicampur dengan  0,05M akan terbentuk larutan buffer dengan pH=4. a) Jika larutan buffer yang terbentuk ditambahkan 0,5mmol  tentukan pHnya

Pertanyaan

60mL larutan begin mathsize 14px style C H subscript 3 C O O H end style 0,2M yang mempunyai pH=3-log 2 dicampur dengan begin mathsize 14px style Na O H end style 0,05M akan terbentuk larutan buffer dengan pH=4.

a) Jika larutan buffer yang terbentuk ditambahkan 0,5mmol begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 end style tentukan pHnyaundefined 

Pembahasan Soal:

Langkah pertama adalah menentukan harga Ka dari asam lemah


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 3 minus sign log space 2 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 3 end exponent end cell 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 open square brackets Asam close square brackets end root end cell row cell 2 cross times 10 to the power of negative sign 3 end exponent end cell equals cell square root of K subscript a cross times 0 comma 2 M end root end cell row cell 4 cross times 10 to the power of negative sign 6 end exponent end cell equals cell K subscript a cross times 0 comma 2 M end cell row cell K subscript a end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 


Kemudian ditentukan mol garam yang terbentuk dari buffer tersebut dari nilai pH buffer tersebut


begin mathsize 14px style 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 1 cross times 10 to the power of negative sign 4 end exponent 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 asam over denominator mol space garam end fraction end cell row cell 10 to the power of negative sign 4 end exponent end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 0 comma 2 M cross times 60 mL over denominator mol space garam end fraction end cell row cell mol space garam end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 0 comma 2 M cross times 60 mL over denominator 10 to the power of negative sign 4 end exponent end fraction end cell row blank equals cell 2 comma 4 mmol end cell end table end style 


Kemudian dapat kita tentukan berapakah pH larutan buffer setelah penambahan 0,5mmol begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 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 K subscript a cross times fraction numerator mol space asam bond mol space basa over denominator mol space garam and mol space basa end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 12 mmol minus sign 0 comma 5 mmol over denominator 2 comma 4 mmol plus 0 comma 5 m mol end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 11 comma 5 mmol over denominator 2 comma 9 mmol end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times 3 comma 97 end cell row blank equals cell 7 comma 93 cross times 10 to the power of negative sign 5 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 7 comma 93 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 7 comma 93 end cell end table end style 


Jadi, pH setelah penambahan 0,5mmol begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 end style adalah 5-log 7,93.undefined 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 April 2021

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