What is the concentration of HCl in a 250.0 mL sample of hydrochloric acid if 15.5 mL of 0.0100 M NaOH is needed to react with all the HCl?
The equation shows the reaction to b 1:1; therefore, the dilution formula may be used.
HCl + NaOH ==> NaCl + H2O
mLsoln1 x Msoln1 = mLsoln2 x Msoln2.
To calculate the concentration of HCl in the 250.0 mL sample of hydrochloric acid, we can use the concept of stoichiometry and the equation of the reaction between HCl and NaOH.
The balanced equation for the reaction is:
HCl + NaOH → NaCl + H2O
From the equation, we can see that one mole of HCl reacts with one mole of NaOH.
We know the volume and concentration of NaOH used (15.5 mL and 0.0100 M, respectively), so we can calculate the number of moles of NaOH used:
moles of NaOH = volume of NaOH (in L) * concentration of NaOH (in mol/L)
= 0.0155 L * 0.0100 mol/L
= 0.000155 mol
Since the reaction is 1:1, the number of moles of HCl in the hydrochloric acid sample is also 0.000155 mol.
Now, convert the volume of the hydrochloric acid sample to liters:
volume of HCl (in L) = 250.0 mL * (1 L / 1000 mL)
= 0.250 L
Finally, calculate the concentration of HCl in the hydrochloric acid sample:
concentration of HCl (in mol/L) = moles of HCl / volume of HCl (in L)
= 0.000155 mol / 0.250 L
Therefore, the concentration of HCl in the 250.0 mL sample of hydrochloric acid is approximately 0.00062 M.