Pertanyaan

Ke d a l am 1 d m 3 l a r u tan ba s a l e mah N H 3 0 , 1 M ( K b = 1 0 − 5 ) d i t ambahkan 1 d m 3 H Cl 0 , 1 M m e n u r u t p ers amaan re ak s i se ba g ai b er ik u t : N H 3 ( a q ) + H Cl ( a q ) → N H 4 Cl ( a q ) B es a r n y a p H l a r u tan se t e l ah d i c am p u r a d a l ah .. ( K w = 1 0 − 14 )

$Ke d a l am 1 d m^{3} l a r u tan ba s a l e mah N H_{3} 0, 1 M (K b = 1 0^{- 5}) d i t ambahkan 1 d m^{3} H Cl 0, 1 M m e n u r u t p ers amaan re ak s i se ba g ai b er ik u t : N H_{3} (a q) + H Cl (a q) \to N H_{4} Cl (a q) B es a r n y a p H l a r u tan se t e l ah d i c am p u r a d a l ah .. (K w = 1 0^{- 14})$

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A. Ratna

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Mahasiswa/Alumni Universitas Negeri Malang

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$space space space space space space space space space space space space space space space space space space space space N H subscript 3 space space space space space space space space space space plus space space space space space H C l space space space space space space space space space space rightwards arrow space space space space N H subscript 4 C l M u l a space space space space space space space space 0 comma 1 space m o l space space space space space space space space space space space space space space space 0 comma 1 space m o l space space space space space space space space R e a k s i space minus 0 comma 1 space m o l space space space space space space space space space space space minus 0 comma 1 space m o l space space space space space space space space space space plus 0 comma 1 space m o l S i s a space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space space space space space space space space bold 0 bold comma bold 1 bold space bold italic m bold italic o bold italic l left square bracket N H subscript 4 C l right square bracket space equals space fraction numerator m o l over denominator V t o t a l end fraction space equals space fraction numerator 0 comma 1 space m o l over denominator 2 space L end fraction space equals space 0 comma 05 space M N H subscript 4 C l space m e r u p a k a n space g a r a m space a s a m comma space s e h i n g g a colon left square bracket H to the power of plus right square bracket space equals space square root of fraction numerator K w space x space left square bracket g a r a m right square bracket over denominator K b end fraction end root space equals space square root of fraction numerator 10 to the power of negative 14 end exponent space x space left square bracket 0 comma 05 right square bracket over denominator 10 to the power of negative 5 end exponent end fraction end root space equals space square root of 5 space x space 10 to the power of negative 11 end exponent end root space equals space square root of 5 space x space 10 to the power of negative 5 comma 5 end exponent p H space equals space minus square root of 5 space x space 10 to the power of negative 5 comma 5 end exponent space M equals space 5 comma 5 space minus space log space square root of 5$

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