How to calculate the molar concentration of the given:

Question

The reaction of 𝐾𝑂𝐻 with 𝐻𝑁𝑂₃ is presented by the unbalanced chemical reaction

below:

___ KOH (aq) + ___ HNO₃ (aq) ---> ___ KNO₃ (aq) + ___ H₂O(l)

Calculate the molar concentration of Nitric acid (𝐻𝑁𝑂₃) when 40.5 − 𝑚𝐿 of 0.674 𝑀 𝐾𝑂𝐻 is required to quench 101.0 𝑚𝐿 of Nitric acid over phenolphthalein indicator.

Answer 1

They don't ask us to, but let's balance the equation first since it's so easy. All coefficients are 1, so it's just:

KOH + HNO₃ → KNO₃ + H₂O

Since we're dealing with a strong acid and strong base, we can assume that K_eq is very large. We know that neutralization (or quenching) occurs when

# mols base = # mols acid

So we have to determine how many moles of base was used. To calculate, we use the equation

M = mols/L

Plug in our values to get

0.674 M* 0.0405 L = 0.0273 mols KOH

Since our molar equivalents in our balanced equation is 1:1, we can assume there were also 0.0273 mols of HNO₃ in our solution. We can use the same molarity equation to determine the concentration of nitric acid since we know moles and L (given in the problem):

M = 0.0273 / 0.101 = 0.27 M HNO₃