Titration Question - How many mL of 0.0350M NaOH are required to titrate 40.0 mL of 0.0350 M HNO3 solution to its equivalence point?

1 Answer
Apr 12, 2018

40ml

Explanation:

There is a shortcut to the answer which I will include at the end, but this is the "long way around".

Both species are strong, i.e both the Nitric acid and the Sodium hydroxide will completely dissociate in aqueous solution.

For a "Strong-Strong" titration, the equivalence point will be exactly at pH=7. (Although Sulfuric acid might be an exception to this as is classified as diprotic in some problems). Anyhow, Nitric acid is monoprotic.

They react in a 1:1 ratio:

NaOH(aq) + HNO_3(aq) -> H_2O(l)+ NaNO_3 (aq)

So to get to the equivalence point an equal amount of mol of HNO_3 must be reacted with NaOH.

Using the concentration formula we can find the moles of HNO_3 in the solution:

c=(n)/v

0.035=(n)/0.04
(40 ml=0.04 liters=0.04 dm^3)

n=0.0014 mol
So we need an equal amount of moles of NaOH, and the titre solution has the same concentration:

0.035=(0.0014)/(v)

v=0.04 dm^3=40ml

Shortcut:
Because we know they react in a 1:1 ratio, and they are of equal concentration, the volume needed to neutralize a solution 40ml of a given concentration HNO_3 will require an equal amount of the same concentration of a solution of NaOH. This leads us right to the answer: 40ml.