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L of NaOH

Someshwar

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7 years ago

What is the pH of solution containing 0.12 mol/L NH4Cl and 0.03 mol/L of NaOH

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bio_man

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#1

Educator

7 years ago

Hi

NH4+Cl- + NaOH ---> NH3 + H2O + NaCl

0.12 0.03 0 initial

0.09 0 0.03 final

it forms bufferr.

pH= pKa + log {[NH3]/ [NH4Cl]}

pH = 9 .25 + log (0.03/0.09)

=9.25 + (-0.48)

= 8.77

Answer: 8.77

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