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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 )

  1. 5 comma 0 space minus space log space 1

  2. 5 comma 5 space minus space log space square root of 5

  3. 7 comma 0 space plus space log space 1

  4. 5 comma 5 space minus space log space square root of 2

  5. 9 comma 0 space plus space log space 1

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A. Ratna

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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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Diketahui tetapan ionisasi basa ( K b ​ ) NH 3 ​ adalah 2 x 1 0 − 5 . pH larutan ( NH 4 ​ ) 2 ​ SO 4 ​ 0 , 1 M adalah ....

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