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Pertanyaan

Jika 27,2 gram K H subscript 2 P O subscript 4 (Mr=136) dilarutkan ke dalam 500 mL air, pH larutan yang terjadi adalah ........

left parenthesis d i k e t a h u i space K a subscript 1 space H subscript 3 P O subscript 4 equals 10 to the power of negative 3 end exponent comma space space K a subscript 2 space H subscript 3 P O subscript 4 equals 10 to the power of negative 8 end exponent comma space space space K a subscript 3 space H subscript 3 P O subscript 4 equals 10 to the power of negative 13 end exponent right parenthesis

  1. 8 + log 2

  2. 8,5 - log 2

  3. 7 - log 2

  4. 4,5 + log 2

  5. 4,5 - log 2

A. Tiara

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

Menentukan konsentrasi garam

M o l space space K H subscript 2 P O subscript 4 space equals space fraction numerator g r a m over denominator M r space end fraction  M o l space space K H subscript 2 P O subscript 4 space equals space fraction numerator 27 comma 2 space g r a m over denominator 136 space g divided by m o l end fraction  M o l space space K H subscript 2 P O subscript 4 space equals space 0 comma 2 space m o l space    space left square bracket K H subscript 2 P O subscript 4 right square bracket space equals space fraction numerator m o l over denominator v o l u m e space l a r u tan space left parenthesis L right parenthesis end fraction  space left square bracket K H subscript 2 P O subscript 4 right square bracket space equals space fraction numerator 0 comma 2 space m o l over denominator 0 comma 5 space l i t e r end fraction  space left square bracket K H subscript 2 P O subscript 4 right square bracket space equals space 0 comma 4 space M

bold italic K bold italic H subscript bold 2 bold italic P bold italic O subscript bold 4 bold spaceadalah garam basa, sehinga :
Ionisasi :

K H subscript 2 P O subscript 4 space left parenthesis a q right parenthesis space rightwards arrow space K to the power of plus space left parenthesis a q right parenthesis space plus space H subscript 2 P O subscript 4 to the power of – space left parenthesis a q right parenthesis space space space space space  0 comma 4 space M space space space space space space space space space space space space space space space space space space space space bold space bold 0 bold comma bold 4 bold space bold italic M

Ka yang dipakai adalah K a subscript 1 , karena hanya ada 1 ion H to the power of plus yang terlibat.

Hidrolisis :

H subscript 2 P O subscript 4 to the power of – space plus space H subscript 2 O space leftwards arrow over rightwards arrow space H subscript 3 P O subscript 4 space plus space O H to the power of –

left square bracket O H to the power of – right square bracket equals square root of fraction numerator K w over denominator K a end fraction cross times open square brackets H subscript 2 P O subscript 4 superscript minus close square brackets end root  left square bracket O H to the power of – right square bracket equals square root of 10 to the power of negative 14 end exponent over 10 to the power of negative 3 end exponent cross times open square brackets 4.10 to the power of negative 1 end exponent close square brackets end root  left square bracket O H to the power of – right square bracket equals square root of 4 cross times 10 to the power of negative 12 end exponent end root  left square bracket O H to the power of – right square bracket equals 2 cross times 10 to the power of – 6 end exponent space M    p O H space equals space – space log space left square bracket O H to the power of – right square bracket space space space  p O H equals space – space log space 2 space x space 10 to the power of – 6 space end exponent space space space  p O H equals space 6 space – space log space 2 space    p H space equals space 14 space – space p O H space space space space  p H space equals space 14 space – space left parenthesis 6 space – space log space 2 right parenthesis space space space  p H space equals space 8 space plus space log space 2

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