1) An NaOH solution of unknown concentration is titrated with HCl.
51.9 mL of 0.400 mol/L HCl will neutralize 75.0 mL of the NaOH.
What is the concentration of the NaOH solution ? Record your answer in decimal notation.
I think the answer is 0.277 mol/L. is that right?
2)An NaOH solution of unknown concentration is titrated with HCl.
33.0 mL of 0.250 mol/L HCl will neutralize 50.0 mL of the NaOH.What is the concentration of the NaOH solution ?Record your answer in decimal notation.
I think the answer is 0.165 mol/L. is that right?
3) A basic solution of known concentration is titrated with a strong acid.
28.5 mL of 0.35 mol/L acid is used to neutralize a 0.30 mol/L base.What volume, in millilitres, of base will be neutralized ?Record your answer in decimal notation.
I am unsure of this one. Please help!
4) A basic solution of known concentration is titrated with a strong acid.
24.1 mL of 0.30 mol/L acid is used to neutralize a 0.10 mol/L base.What volume, in millilitres, of base will be neutralized ?Record your answer in decimal notation. Also unsure of this one. Please Help!
1 and 2 are correct.
#3. Technically, the problem SHOULD give you enough information to know if this is a 1:1 reaction or some other kind. It doesn't so I assume we assume it is 1:1; i.e., HA + BOH ==> BA + H2O
moles acid = M x L = ?
moles base = moles acid
M base = moles/L soln
You have M and moles, solve for L and convert to mL.
#4.
See my note for #3. This one is worked the same as #3 and we must make the same assumption.
I still don't understand. Capital M x L
M = molarity
L = liter
molarity is moles/L; therefore, moles = M x L.
1) To find the concentration of the NaOH solution, we need to use the equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
First, we can calculate the number of moles of HCl used in the titration:
51.9 mL of 0.400 mol/L HCl = (51.9 mL) x (0.400 mol/L) = 20.76 mmol
Since the balanced equation tells us that the stoichiometry of NaOH to HCl is 1:1, the number of moles of NaOH used in the titration is also 20.76 mmol.
Next, we can calculate the concentration of the NaOH solution:
75.0 mL of NaOH = 20.76 mmol of NaOH
Concentration = (20.76 mmol) / (75.0 mL) = 0.2768 mol/L
Therefore, the concentration of the NaOH solution is approximately 0.277 mol/L.
Your answer of 0.277 mol/L is correct.
2) Following the same steps as above:
33.0 mL of 0.250 mol/L HCl = (33.0 mL) x (0.250 mol/L) = 8.25 mmol
The number of moles of NaOH used in the titration is also 8.25 mmol.
50.0 mL of NaOH = 8.25 mmol of NaOH
Concentration = (8.25 mmol) / (50.0 mL) = 0.165 mol/L
Therefore, the concentration of the NaOH solution is indeed 0.165 mol/L.
Your answer is correct.
3) In this case, we have a known concentration of the acid and need to find the volume of base required for neutralization.
Using the equation for the reaction:
Base + Acid -> Salt + H2O
The stoichiometry of the acid to the base is 1:1. So, the reaction requires equal numbers of moles of acid and base for neutralization.
Given that 28.5 mL of 0.35 mol/L acid is used, we can calculate the number of moles of acid:
(28.5 mL) x (0.35 mol/L) = 9.975 mmol
Since the stoichiometry is 1:1, this means that 9.975 mmol of base will be neutralized.
To find the volume of base required, we need to know its concentration. The question mentions a 0.30 mol/L base concentration, so the volume of base can be calculated as:
Volume = (9.975 mmol) / (0.30 mol/L) = 33.25 mL
Therefore, the volume of base to be neutralized is approximately 33.25 mL.
4) Following a similar approach:
Given that 24.1 mL of 0.30 mol/L acid is used, we can calculate the number of moles of acid:
(24.1 mL) x (0.30 mol/L) = 7.23 mmol
Since the stoichiometry is 1:1, this means that 7.23 mmol of base will be neutralized.
Using the concentration of the base (0.10 mol/L), we can calculate the volume of base required:
Volume = (7.23 mmol) / (0.10 mol/L) = 72.3 mL
Therefore, the volume of base to be neutralized is approximately 72.3 mL.
I hope this helps!