In: Chemistry
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?
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