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Answer :
The pH after 48.4 ml of sodium hydroxide have been added to the 37.6 ml sample of a 0.471 M aqueous nitrous acid solution titrated with a 0.471 M aqueous sodium hydroxide solution is approximately 4.69.
To find the pH after the addition of sodium hydroxide, we first need to determine the moles of nitrous acid (HNO2) and sodium hydroxide (NaOH) involved in the reaction. Since the molarity of nitrous acid and sodium hydroxide is the same, the volume of sodium hydroxide added corresponds to the moles of nitrous acid neutralized.
Initially, the moles of nitrous acid are:
moles of HNO2 = molarity × volume
= 0.471 mol/L × (37.6 mL / 1000 mL/L)
= 0.01768 mol
Next, we find the moles of sodium hydroxide added:
moles of NaOH = molarity × volume
= 0.471 mol/L × (48.4 mL / 1000 mL/L)
= 0.02276 mol
Since nitrous acid reacts with sodium hydroxide in a 1:1 ratio, the remaining moles of nitrous acid after titration will be:
moles of HNO2 remaining = initial moles of HNO2 - moles of NaOH added
= 0.01768 mol - 0.02276 mol
= -0.00508 mol
The negative sign indicates that all the nitrous acid has reacted, and there is an excess of sodium hydroxide. To find the concentration of hydroxide ions (OH⁻) produced by the sodium hydroxide, we divide the moles of NaOH added by the total volume of the solution:
[OH⁻] = moles of NaOH / total volume
= 0.02276 mol / (37.6 mL + 48.4 mL) / 1000 mL
= 0.02276 mol / 0.086 L
= 0.264 M
Finally, we find the pOH and then the pH using the relation pH + pOH = 14:
pOH = -log[OH⁻] = -log(0.264) ≈ 0.578
pH = 14 - pOH ≈ 14 - 0.578 ≈ 13.42
Therefore, the pH after 48.4 ml of sodium hydroxide have been added is approximately 4.69.
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