Henry uses 0.150L of 0.500 mol/L NaOH to neutralize 250.0 mL of the hydrochloric acid. What is the concentration of the HCl (aq)?
method 1:
NaOH + HCl ==> NaCl + H2O
mols NaOH = M x L = ?
mols HCl = mols HCl (look at the equation; note 1 mol NaOH = 1 mol HCl)
Then M HCl = mols HCl/L HCl = ?
method 2:
NaOH + HCl ==> NaCl + H2O
Note 1 mol NaOH = 1 mol HCl
mL NaOH x M NaOH = mL HCl x M HCl
Substitute and solve for M HCl.
To find the concentration of HCl (aq), we can use the concept of the mole ratio and the equation:
Moles of Substance A = Moles of Substance B
In this case, Substance A is NaOH and Substance B is HCl.
First, let's calculate the number of moles of NaOH used:
Moles of NaOH = Volume of NaOH (in liters) × Concentration of NaOH (in mol/L)
= 0.150 L × 0.500 mol/L
= 0.075 mol
Since the equation is balanced as 1:1, the number of moles of HCl is also equal to 0.075 mol.
Now we need to calculate the concentration of HCl. We have the volume of HCl (250.0 mL) given in milliliters, so we need to convert it to liters by dividing it by 1000:
Volume of HCl (in liters) = 250.0 mL ÷ 1000
= 0.250 L
Finally, we can calculate the concentration of HCl:
Concentration of HCl (aq) = Moles of HCl / Volume of HCl (in liters)
= 0.075 mol / 0.250 L
= 0.300 mol/L
Therefore, the concentration of the HCl (aq) is 0.300 mol/L.