25.0 mL of an unknown HCl solution is titrated with 17.5 mL 0.100 M NaOH. What is the
Molarity of the HCl solution?
titration equation:
Ma*Va=Mb*Vb
Ma=.1*17.5ml/25.0
To find the molarity of the HCl solution, we can use the equation:
M1V1 = M2V2
where M1 is the molarity of the NaOH solution, V1 is the volume of the NaOH solution used, M2 is the molarity of the HCl solution, and V2 is the volume of the HCl solution.
Plugging in the given values:
M1 = 0.100 M
V1 = 17.5 mL
V2 = 25.0 mL
M1V1 = M2V2
(0.100 M)(17.5 mL) = M2(25.0 mL)
Simplifying the equation:
1.75 M/mL = M2(25.0 mL)
Dividing both sides by 25.0 mL:
1.75 M/mL ÷ 25.0 mL = M2
M2 = 0.070 M
Therefore, the molarity of the HCl solution is 0.070 M.
To find the molarity of the HCl solution, you can use the concept of stoichiometry and the balanced equation for the reaction between HCl and NaOH.
The balanced equation for the reaction is:
HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)
From the balanced equation, you can see that the stoichiometric ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, there is 1 mole of NaOH.
Given that the volume of NaOH used is 17.5 mL and the concentration of NaOH is 0.100 M, you can calculate the number of moles of NaOH used:
moles of NaOH = volume of NaOH (in L) * concentration of NaOH
= 17.5 mL * (1 L / 1000 mL) * 0.100 mol/L
= 0.00175 mol
Since the stoichiometric ratio between HCl and NaOH is 1:1, the number of moles of HCl is also 0.00175 mol.
Now, you need to find the molarity of the HCl solution. Molarity is defined as the number of moles of solute divided by the volume of solution in liters. In this case, the volume of solution is given as 25.0 mL, which is equivalent to 0.025 L.
Molarity of HCl solution = moles of HCl / volume of solution
= 0.00175 mol / 0.025 L
= 0.07 M
Therefore, the molarity of the HCl solution is 0.07 M.