By titration it is found that 20.5 mL of 0.168 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution
HCl + NaOH ==> NaCl + H2O
mols NaOH = M x L =?
Using the coefficients in the balanced equation, convert mols NaOH to mols HCl.
Then M HCl = mols HCl/L HCl.
Solve for M HCl.
To calculate the concentration of the HCl solution, we can use the concept of stoichiometry and the equation for the neutralization reaction between NaOH and HCl:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
From the balanced equation, we can see that the molar ratio between NaOH and HCl is 1:1. This means that for every 1 mole of NaOH, we will need 1 mole of HCl to completely neutralize it.
Given that the volume of NaOH used is 20.5 mL and the concentration of NaOH is 0.168 M, we can calculate the number of moles of NaOH used:
moles of NaOH = concentration of NaOH × volume of NaOH
= 0.168 mol/L × 0.0205 L
= 0.003444 mol
Since the molar ratio between NaOH and HCl is 1:1, the number of moles of HCl required to neutralize the NaOH is also 0.003444 mol.
Now, we can calculate the concentration of HCl using the volume of HCl and the number of moles of HCl:
concentration of HCl = moles of HCl / volume of HCl
= 0.003444 mol / 0.0250 L
= 0.13776 mol/L
Rounding to the appropriate number of significant figures, the concentration of the HCl solution is 0.138 M.