A student successfully neutralizes 25.0mL of HCl(aq) during a titration experiment with 18.5mL of 0.150M NaOH(aq). What is the c
1 answer:
Answer:
The concentration is 0.111 M.
Explanation:
Use dimensional analysis to solve the problem. Since both HCl and NaOH are strong, the dissociate completely in solution.
For this neutralization, the reaction is HCl + NaOH = NaCl + HOH
18.5 mL x x
x
= 0.002775 moles HCl
Molarity = moles/ Liters
Molarity of HCl = 0.002775 moles HCl / 0.025 L = 0.111 M
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Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
First we need to calculate the number of moles of NaOH titrated.
molar concentration = number of moles / solution volume (liter)
number of moles = molar concentration × solution volume
number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles
Then we look at the chemical reaction:
CH-COOH + NaOH = CH
COONa + H
O
We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.
Now we can get the mass of acetic acid:
number of moles = mass (grams) / molecular mass (g/mol)
mass = number of moles × molecular mass
mass of acetic acid = 0,01125 × 60 = 0.675 g
We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:
concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100
concentration of acetic acid = (0.674 / 7) × 100 = 9.6 %