Please tell me if these are right!
1.How many milliliters of 0.215 M NaOH solution are needed to completely neutralize 2.50 ml of 0.825 M H2SO4 solution?
I got 19.2 ml
2. A 10.0-ml sample of vinegar which is an aqueous solution of acetic acid HC2G3O2 requires 16.5 ml of 0.500M NaOH to reach the endpoint in a titration what is the morality of the acetic acid solution?
I got 0.00825 M acetic acid solution
thank you
1. correct
2. acetic acid reads CH3COOH when I was in school
10.0*M = 16.5*0.500
I get M=0.825
To verify the answers, let's go through the calculations for each question.
1. How many milliliters of 0.215 M NaOH solution are needed to completely neutralize 2.50 ml of 0.825 M H2SO4 solution?
To solve this problem, you need to use the concept of stoichiometry and the balanced equation for the reaction between NaOH and H2SO4. The balanced equation is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
The stoichiometry of the reaction tells us that for every 2 moles of NaOH, we need 1 mole of H2SO4.
First, let's calculate the number of moles of H2SO4 in 2.50 ml of 0.825 M solution:
Moles of H2SO4 = volume (in liters) x concentration (in mol/L)
= 2.50 ml x (1 L/1000 ml) x 0.825 mol/L
= 0.00206 mol
According to the stoichiometry, we need twice as many moles of NaOH. Therefore, we need:
Moles of NaOH = 2 x Moles of H2SO4
= 2 x 0.00206 mol
= 0.00412 mol
Now, let's calculate the volume of 0.215 M NaOH solution needed to reach 0.00412 moles:
Volume of NaOH = Moles of NaOH / Concentration of NaOH
= 0.00412 mol / 0.215 mol/L
= 0.0191 L
Finally, convert the volume to milliliters:
Volume of NaOH = 0.0191 L x (1000 ml/1 L)
= 19.1 ml
Therefore, the correct answer is 19.1 ml, not 19.2 ml.
2. A 10.0-ml sample of vinegar, which is an aqueous solution of acetic acid (HC2H3O2), requires 16.5 ml of 0.500 M NaOH to reach the endpoint in a titration. What is the molarity of the acetic acid solution?
In this question, we can use the balanced equation for the reaction between acetic acid and NaOH:
HC2H3O2 + NaOH -> NaC2H3O2 + H2O
The stoichiometry of the reaction tells us that for every 1 mole of acetic acid, we need 1 mole of NaOH.
Since the volumes of the vinegar and NaOH used in the titration are in a 1:1 ratio, we can calculate the moles of acetic acid:
Moles of HC2H3O2 = Volume of NaOH x Concentration of NaOH
= 16.5 ml x (1 L/1000 ml) x 0.500 mol/L
= 0.00825 mol
The moles of acetic acid can be equated to the molarity of the acetic acid solution (M):
Moles of HC2H3O2 = M x Volume of acetic acid (in L)
0.00825 mol = M x (10.0 ml x (1 L/1000 ml))
M = 0.00825 mol / 0.0100 L
M = 0.825 M
Therefore, the molarity of the acetic acid solution is 0.825 M, not 0.00825 M.
So, the correct answers are:
1. 19.1 ml
2. 0.825 M