Question

A 25.00 −ml sample of an unknown hclo4 solution requires titration with 21.52 ml of 0.2000 m naoh to reach the equivalence point. part a what is the concentration of the unknown hclo4 solution? the neutralization reaction is: hclo4(aq)+naoh(aq)→h2o(l)+naclo4(aq)

Answer 1

The concentration of the unknown HClO₄ solution: 0.17216  m

Further explanation

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint / equivalence point of the reaction can use indicators according to the appropriate pH range

Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.

An acid-base titration there will be a change in the pH of the solution.

From this pH change a Titration Curve can be made which is a function of acid / base volume and pH of the solution

Acid-base titration formula

Ma Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

Neutralization Reaction:

HClO₄ (aq) + NaOH (aq) → H₂O (l) + NaClO₄ (aq)

21.52 ml of 0.2000 m NaOH

25.00 ml HClO₄ solution

Acid-base titration formula

Ma Va. na = Mb. Vb. nb

a = NaOH, b = HClO₄ (both have valence 1)

0.20 m. 21.52. 1 = Mb. 25 1

$\rm Mb = \dfrac {0.20 \times 21.52} {25} \\\\ Mb = \boxed {\bold {0.17216 \: m}}$

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Answer 2

The neutralization reaction is;
HClO₄(aq) + NaOH(aq) --> H₂O(l) + NaClO₄(aq)
Stoichiometry of HClO₄ to NaOH is 1:1
Therefore at equivalence an equal amount of NaOH reacts with an equal amount of HClO₄
The number of moles of NaOH in 1 L - 0.2 mol
Therefore in 21.52 ml of NaOH - 0.2/1000 x 21.52 = 0.004304 mol
the amount of HClO₄ moles that have reacted - 0.004304  mol
In 25 ml of HClO₄ - 0.004304 mol
Therefore in 1000 ml of HClO₄ - 0.004304/25 x 1000 = 0.172 mol
concentration of HClO₄ - 0.172 mol/L