Titration of 25.00 mL of a monoprotic acid solution with 0.2269 M NaOH is:
Initial NaOH = 3.96 mL
Final NaOH = 38.84 mL (dif is 34.88mL)
Based on data what is the molar concentration of acid?
I did
V x M = 0.03488L NaOH x 0.2269 = 0.079143
Not sure if I should have switched to Liters.
Then M1 x V2 = M2 x V2
M Mor Acid x 0.025 L = 0.2269M NaOH x 0.03488L = 0.07914 / 0.025 = 3.1656
Is this correct or should I have left them in mL
Actually it makes no difference as long as you keep the units straight. However, I think you made a math error of about 10 tims.
0.03488 x 0.2269 = 0.007914 mol NaOH.
mols acid = mols NaOH
M acid = mols acid/L acid = 0.007914/0.025 = aboaut 0.32M but you need to redo, do it more accurately, and answer with the right number of significant figures.
If you to do it in mL, it is done this way.
34.88 x 0.2269M = 7.914 millimoles.
Then M = mmoles/mL = 7.914/25 = about 0.32 M.
Oops. Thanks. I see I copied 0.007943 incorrectly. I'll be more careful in the future.
Your approach to solving the problem is correct. However, there are a couple of mistakes in your calculations.
First, let's correct your initial calculation:
Initial NaOH = 3.96 mL
Final NaOH = 38.84 mL (difference is 34.88 mL)
To calculate the amount of NaOH used in the titration, you need to subtract the initial NaOH volume from the final NaOH volume:
NaOH used = Final NaOH - Initial NaOH
NaOH used = 38.84 mL - 3.96 mL = 34.88 mL = 0.03488 L (since 1 L = 1000 mL)
Now, let's calculate the molarity of the acid:
M1 x V1 = M2 x V2
where:
M1 = molarity of the acid
V1 = volume of the acid solution in liters
M2 = molarity of NaOH
V2 = volume of NaOH used in liters
We know that:
V1 = 25.00 mL = 0.025 L (since 1 L = 1000 mL)
M2 = 0.2269 M (given)
Now, let's plug in the values and solve for M1:
M1 x 0.025 L = 0.2269 M x 0.03488 L
M1 = (0.2269 M x 0.03488 L) / 0.025 L
M1 = 0.3136 M
Therefore, the molar concentration of the acid is 0.3136 M.
To summarize, you should have switched to liters when performing the calculations, and the correct molar concentration of the acid is 0.3136 M.