Question

References Use the References to access important values if needed for this question. An aqueous solution contains 0.456 M nitrous acid. How many mL of 0.3 10 M sodium hydroxide would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 3.290? mL Submit Answer

Answer 1

Answer:

Addition of aq.NaOH to HNO₂ gives NaNO₂. This will form NaNO₂-HNO₂ buffer. pKa of HNO₂ = 3.15.

The pertinent equation is,

HNO₂ (aq) + NaOH (aq) $\rightarrow$ NaNO₂ (aq) + H₂O.

Its clear from mole relationship that, # of moles of NaNO2 formed = # of moles of NaOH added.

pH of the buffer is given by Henderson-Hasselbalch equation as,

pH = pKa + log([Base or salt]/[Acid]) or

pH = pKa + log(# of moles of base or salt/# of moles of Acid)

pH = pKa + log(# of moles of NaNO₂/# of moles of HNO₂) .......... (1)

Given: pH = 3.290, Initial Molarity of HNO2 = 0.456 M, Volume = 125.0 mL = 0.125 L

So, # of moles of HNO₂ initially = 0.456 * 0.125 = 0.057 moles.

Let, # of moles of NaNO₂ formed at equilibrium after addition of NaOH = A then

# of moles of HNO2 left at equilibrium = (0.057-A) moles

Placing all known data in eq.1

3.290 = 3.150 + log[A/(0.057-A)]

3.290-3.150 =  log[A/(0.057-A)]

log[A/(0.057-A)] = 0.14

A/(0.057-A) = 10^0.14.

A/(0.057-A) = 1.38

A = 1.38*(0.057-A)

A + 1.38A = 1.38*0.057

2.38A = 1.38*0.057

A = 1.38*0.057 / 2.38

A = 0.033.

I.e. # of moles of NaNO2 formed =0.033 mol

I.e. # of moles of NaOH formed = 0.033 mol

Now,

Molarity of NaOH = 0.310 M = 0.310 mol/L, Volume of NaOH = ?

We know that,

# of moles of NaOH = Molarity * Volume

Placing known data,

0.033 mol = 0.310 mol/L * Volume

Volume = 0.033 mol / 0.310 mol/L

Volume = 0.1066 L = 106.6 mL

Volume of NaOH = 106.6 mL

106.6 mL of 0.310 M NaOH need to be added to 125.0 mL of 0.456 M HNO2 to have buffer with pH 3.290.

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