Chemistry, 24.10.2019 09:43 nakarelinp0p303

When titrating a strong monoprotic acid and koh at 25°c, the ph will be equal too?

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The pH will be equal to $\boxed{\text{7}}$ in the titration of a strong monoprotic acid and KOH at ${\text{25}}^\circ{\text{C}}$ .

Further Explanation:

Neutralization reaction:

It is the reaction that occurs between an acid and a base in order to form salt and water. It is named so as it neutralizes the excess amount of hydrogen or hydroxide ions present in the solution. It is used to decrease the acidity in the stomach, wastewater treatment, antacid tablets, and to control the pH of soil. An example of a neutralization reaction is,

${\text{HCl}}+{\text{NaOH}}\to{\text{NaCl}}+{{\text{H}}_2}{\text{O}}$

Titration:

It is a laboratory technique to determine the concentration of an unknown solution with the help of a known concentration. The known solution (titrant) is added from burette to a known quantity of unknown solution (analyte).

Consider HA to be a strong monoprotic acid. KOH is also a strong base. The titration between HA and KOH would result in the following neutralization reaction:

${\text{HA}}+{\text{KOH}}\to{\text{KA}}+{{\text{H}}_{\text{2}}}{\text{O}}$

Since both acids and bases are strong, complete neutralization occurs and salt and water are formed as products. The resultant solution will be neutral and therefore its pH at ${\text{25}}^\circ{\text{C}}$ will be equal to 7.

Learn more:

1. Name the salts produced in each of the neutralization reactions:

2. Determine the moles of water produced:

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Acid-base reactions

Keywords: Acids, bases, salt, water, KOH, HA, H2O, KA, titration, technique, analyte, titrant, known, unknown, neutralization reaction, hydrogen, hydroxide ion, strong monoprotic acid, pH, 7, neutral.

Answer from: Quest

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Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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