Chemistry, 13.12.2020 23:10 rm3582317

Student carries out a titration to determine the concentration of a solution of nitric acid. She titrates the solution of nitric acid against a standard solution
of sodium hydroxide with a known concentration of 0.0998 mol/dm². She
finds that 21.80 cm of the nitric acid solution is needed to exactly neutralise
25.0 cm of the sodium hydroxide solution.
Calculate the concentration of the nitric acid solution. Give your answer to
three significant figures.
The equation for the neutralisation reaction is
HNO3 + NaOH → NaNO3 + H2O

Answers: 3

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Answers

Answer from: fazegaze89

The concentration of the nitric acid solution is 0.11445 mol/dm³

From the question,

We are to calculate the concentration of the nitric acid solution

The given balanced chemical equation for the reaction is

HNO₃ + NaOH → NaNO₃ + H₂O

This means 1 mole of HNO₃ is needed to completely neutralize 1 mole of NaOH

Using the formula

$\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}$

Where $C_{A}$ is the concentration of acid

$C_{B}$ is the concentration of base

$V_{A}$ is the volume of acid

$V_{B}$ is the volume of base

$n_{A}$ is the mole ratio of acid

$n_{B}$ is the mole ratio of base

From the given information

$C_{B}= 0.0998\ mol/dm^{3}$

$V_{A} = 21.80 \ cm^{3}$

$V_{B} = 25.0 \ cm^{3}$

From the balanced chemical equation

$n_{A} = 1$

$n_{B} =1$

Putting the parameters into the formula, we get

$\frac{C_{A} \times 21.80 }{0.0998 \times 25.0} = \frac{1}{1}$

Then,

$C_{A} \times 21.80=0.0998 \times 25.0$

∴ $C_{A}=\frac{0.0998 \times 25.0}{21.80}$

$C_{A} =\frac{2.495}{21.80}$

$C_{A} = 0.11445 \ mol/dm^{3}$

Hence, the concentration of the nitric acid solution is 0.11445 mol/dm³

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Answer from: queenfan1818

The concentration of the Nitric acid solution : 0.114 M

Further explanation

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

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

Reaction

HNO₃ + NaOH → NaNO₃ + H₂O

Concentration a standard solution of sodium hydroxide : 0.0998 mol/dm³ , and the volume = 25 cm³

moles NaOH=

$\tt mol=M\times V\\\\mol=0.0998\times 25\\\\mol=2.495~mlmoles$

From the equation, mol ratio HNO₃ : NaOH = 1 : 1, so mol HNO₃ = mol NaOH=2.495 mlmoles

The volume of HNO₃ = 21.8 cm³, so the concentration :

$\tt M=\dfrac{n}{V}\\\\M=\dfrac{2.495}{21.8}\\\\M=0.114$

Answer from: Quest

it should be all of them because it asks if you included all of those in your answer

explanation:

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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