If you could help me it would be greatly appreciated, Thanx
Question 1
In a titration, a 20mL sampl of NaOh(aq) was neutralized by 14.9 mL of 0.13 mol/L H2SO4(aq). The concentration of the base is
A. 8.7 mol/L
B. 0.35 mol/L
C. 0.19 mol/L
D. 0.048 mol/L
Question 2
A lab technician diluted a 25.0 mL sample of ethanoic acid, with an initial concentration of 18.0 mol/L, to obtain a molar concentration of 0.500 mol/L. To what volume of water did the technician add the 25 mL of acid? Show complete calculations.
If you could help me it would be greatly appreciated, Thanx
Question 1
In a titration, a 20mL sampl of NaOh(aq) was neutralized by 14.9 mL of 0.13 mol/L H2SO4(aq). The concentration of the base is
H2SO4 + 2NaOH ==> 2H2O+ Na2SO4
mols H2SO4 = L x M = ??
Use the coefficients in the equation to convert mols H2SO4 to mols NaOH.
mols NaOH = ??mol H2SO4 x (2 mol NaOH/1 mol H2SO4 = xx mol NaOH.
Then use the definition of M = mols/L to convert xx mol NaOH to M (the volume is 0.020 L)
A. 8.7 mol/L
B. 0.35 mol/L
C. 0.19 mol/L
D. 0.048 mol/L
Question 2
A lab technician diluted a 25.0 mL sample of ethanoic acid, with an initial concentration of 18.0 mol/L, to obtain a molar concentration of 0.500 mol/L. To what volume of water did the technician add the 25 mL of acid? Show complete calculations
mL x M = mL x M
Show your work if you get stuck and need further assistance.
Question 1:
To find the concentration of the base, NaOH, in the titration, we can use the stoichiometry of the reaction between NaOH and H2SO4.
The balanced chemical equation for the reaction is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. So, to find the moles of NaOH, we need to first find the moles of H2SO4.
To find the moles of H2SO4, we can use the formula:
moles = volume (in liters) x concentration (in mol/L)
Given:
Volume of H2SO4 = 14.9 mL = 14.9/1000 L = 0.0149 L
Concentration of H2SO4 = 0.13 mol/L
Moles of H2SO4 = 0.0149 L x 0.13 mol/L = 0.001937 mol
Now, we can use the stoichiometry of the reaction to find the moles of NaOH.
moles of NaOH = moles of H2SO4 x (2 mol NaOH / 1 mol H2SO4)
moles of NaOH = 0.001937 mol x (2 mol NaOH / 1 mol H2SO4) = 0.003874 mol
Finally, to find the concentration of NaOH, we divide the moles of NaOH by the volume of the NaOH solution used in the titration:
Concentration of NaOH = moles of NaOH / volume of NaOH solution
Concentration of NaOH = 0.003874 mol / 0.020 L = 0.1937 mol/L
Therefore, the concentration of the base, NaOH, is 0.1937 mol/L.
Question 2:
To find the volume of water added to the 25 mL sample of ethanoic acid, we can use the dilution formula.
The dilution formula is:
M1V1 = M2V2
Where:
M1 = initial molar concentration
V1 = initial volume
M2 = final molar concentration
V2 = final volume
Given:
M1 = 18.0 mol/L
V1 = 25.0 mL = 25/1000 L = 0.025 L
M2 = 0.500 mol/L
We want to find V2, the final volume of the diluted solution.
Rearranging the dilution formula, we have:
V2 = (M1 x V1) / M2
Substituting the given values:
V2 = (18.0 mol/L x 0.025 L) / 0.500 mol/L
V2 = 0.45 L
Therefore, the technician added 0.45 L (or 450 mL) of water to the 25 mL sample of ethanoic acid to obtain a molar concentration of 0.500 mol/L.