if a 20 ml sample of nitric acid is titrated with 32 ml of 0.200 mol of potassium hydroxide what is the concentration of the nitric acid
HNO3 + KOH ==> H2O + KNO3
mols KOH = M x L = ?
Using the coefficients in the balanced equation, convert mols KOH to mols HNO3. In this case mols KOH = mols HNO3.
Then M HNO3 = mols HNO3/L HNO3 = ?
To find the concentration of the nitric acid, we can use the concept of stoichiometry and the equation:
HNO3 + KOH → KNO3 + H2O
Given that 32 ml of 0.200 M potassium hydroxide (KOH) is used in the reaction, we can determine the number of moles of potassium hydroxide:
moles of KOH = volume (L) × concentration (mol/L)
= 0.032 L × 0.200 mol/L
= 0.0064 mol
According to the balanced equation, the stoichiometric ratio between KOH and HNO3 is 1:1. Therefore, the moles of nitric acid (HNO3) used in the reaction is also 0.0064 mol.
Since we are given a 20 ml sample of nitric acid, we first need to calculate the concentration before the reaction:
concentration (mol/L) = moles/volume (L)
= 0.0064 mol / 0.020 L
= 0.32 M
Hence, the concentration of the nitric acid is 0.32 M.