# 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?

Apr 12, 2018

$40 m l$

#### 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 $p H = 7$. (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:

$N a O H \left(a q\right) + H N {O}_{3} \left(a q\right) \to {H}_{2} O \left(l\right) + N a N {O}_{3} \left(a q\right)$

So to get to the equivalence point an equal amount of mol of $H N {O}_{3}$ must be reacted with $N a O H$.

Using the concentration formula we can find the moles of $H N {O}_{3}$ in the solution:

$c = \frac{n}{v}$

$0.035 = \frac{n}{0.04}$
(40 ml=0.04 liters=0.04 dm^3)

$n = 0.0014 m o l$
So we need an equal amount of moles of $N a O H$, and the titre solution has the same concentration:

$0.035 = \frac{0.0014}{v}$

$v = 0.04 {\mathrm{dm}}^{3}$=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 $H N {O}_{3}$ will require an equal amount of the same concentration of a solution of $N a O H$. This leads us right to the answer: $40 m l$.