Question

A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine if each addition would exceed the capacity of the buffer to neutralize it.

a. 250 mg NaOH
b. 350 mg KOH
c. 125 g HBr
d. 1.35g Hl

Answer 1

no one additions exceed the capacity of the buffer

Step-by-step explanation:

given

Volume buffer = 500.0 mL = 0.5 L

mol HNO₂ = 0.5 L × 0.100 mol/L = 0.05 mol HNO₂

mol NO₂⁻ = 0.5 L × 0.150 mol/L = 0.075 mol NO₂⁻

solution

we know when any base more than 0.05 (HNO2) than exceed buffer capacity

and when any base more than 0.075 (KNO2) than exceed buffer capacity

when we add 250 mg NaOH (0.250 g)

than molar mass NaOH =40 g/mol

and mol NaOH = 0.250 g ÷ 40g/mol

mol NaOH = 0.00625 mol

0.00625 mol NaOH will be neutralized by 0.00625 mol HNO₂

so it would not exceed the capacity of the buffer.

and

when we add 350 mg KOH (0.350 g)

than molar mass KOH =56.10 g

and mol KOH = 0.350 g ÷ 56.10 g/mol

mol KOH = 0.0062 mol

here also capacity of the buffer will not be exceeded

and

now we add 1.25 g HBr

than molar mass HBr = 80.91 g/mol

and mol HBr = 1.25 g ÷ 80.91 g/mol

mol HBr = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻

so the capacity will not be exceeded.

and

we add 1.35 g HI

molar mass HI = 127.91 g/mol

so mol HI = 1.35 g ÷ 127.91 g/mol

mol HI = 0.011 mol

capacity of the buffer will not be exceed