If a chemist titrates 120.0 mL of NaOH with a 5.0 M solution of HCl and requires 56.0 mL of the acid to reach the endpoint, what is the concentration of the NaOH?
Same procedure as for you H2SO4/NaOH problem above but start with moles H2SO4, convert to moles NaOH, solve for M NaOH.
5 M
To find the concentration of NaOH, you need to use the equation:
M1V1 = M2V2
Where:
M1 = concentration of NaOH
V1 = volume of NaOH (in liters)
M2 = concentration of HCl
V2 = volume of HCl (in liters)
First, let's convert the volumes to liters:
V1 = 120.0 mL = 0.120 L
V2 = 56.0 mL = 0.056 L
Next, plug in the values into the equation:
M1 * 0.120 L = 5.0 M * 0.056 L
Solving for M1, we get:
M1 = (5.0 M * 0.056 L) / 0.120 L
Simplifying the equation, we have:
M1 = 2.333 M
Therefore, the concentration of NaOH is 2.333 M.
To find the concentration of NaOH, we can use the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
From the equation, we can see that the ratio of NaOH to HCl is 1:1. This means that for every 1 mole of NaOH, we need 1 mole of HCl to reach the endpoint.
Given that the volume of the HCl solution used is 56.0 mL (0.056 L) and its concentration is 5.0 M, we can calculate the number of moles of HCl used:
moles HCl = concentration × volume
moles HCl = 5.0 M × 0.056 L = 0.28 moles
Since NaOH and HCl react in a 1:1 ratio, the number of moles of NaOH used is also 0.28 moles.
Now, let's calculate the concentration of NaOH in the original solution.
We know the volume of the NaOH solution used is 120.0 mL (0.120 L), and the number of moles of NaOH used is 0.28 moles.
Concentration of NaOH = moles NaOH / volume NaOH
Concentration of NaOH = 0.28 moles / 0.120 L
Concentration of NaOH = 2.33 M
Therefore, the concentration of NaOH is 2.33 M.