Question

# What volume (to the nearest 0.1 mL) of 7.40-M NaOH must be added to 0.400 L...

What volume (to the nearest 0.1 mL) of 7.40-M NaOH must be added to 0.400 L of 0.350-M HNO2 to prepare a pH = 3.80 buffer?

For this we have the equation

OH- + HNO2 ---> NO2- + H2O

pKa of HNO2= 3.39 (ref table http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf)

Henderson–Hasselbalch equation:

pH = pKa + log([NO2-]/[HNO2]) concentration of NO2- is substituted for concentration of NaOH as amount of NO2- is dependent on the amount of NaOH.
3.80 = 3.39 + log(7.40x/ (0.400 L x 0.350 M - 7.40x)) Concentration of HNO2 is the amount of HNO2 added subtrated by the amount reacted with NaOH

where x is volume in Liters

3.8 - 3.39 = log(7.40x/ (0.400 L x 0.350 M - 7.40x))

100.41 = 7.40x/ 0.14 - 7.40x
(2.57)(0.14 - 7.40x) = 7.40x
0.359 - 19.02x = 7.40x
0.359 = 7.40x + 19.02x

x = 0.359/26.42 = 0.01358 Litres = 13.58 mL

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