What volume (to the nearest 0.1 mL) of 5.20-M NaOH must be added to 0.500 L...

What volume (to the nearest 0.1 mL) of 5.20-M NaOH must be added to 0.500 L of 0.350-M HNO₂ to prepare a pH = 3.50 buffer?

Solutions

Expert Solution

OH- + HNO2NO2- + H2O

this neutraization reaction happens in the soution,when strong base NaOH is completely used up.

moles of NaOH=5.20 M*V [v=required volume]

moles of HNO2=0.35 mol/L*0.5L=0.17 mol

ICE table

[OH-] [ HNO2] [NO2-]

initial 5.20*V 0.17 0

change -5.20V -0.520V +0.520V

equilibrium 0.0 0.17-0.520V 0.520V

using henderson-hasselbach equation, you can calculate the [Base]/[acid] =[NO2-]/[HNO2] needed to get a pH=3.50 with the buffer. NaNO2/HNO2

pH=pka+log [NO2-]/[HNO2]

pka for NaNO2/HNO2 buffer=3.40

plug in the values,

3.50=3.40+log [NO2-]/[HNO2]

0.10=log [NO2-]/[HNO2]

[NO2-]/[HNO2]=10^0.10=1.26

or,

[NO2-]/[HNO2]=1.26

0.520V/( 0.17-0.520V)=1.26

or,0.520V=0.214-0.655V

1.175V=0.214

V=0.182 L=18.2 ml

volume of NaOH=18.2 ml

queen_honey_blossom answered 1 year ago

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