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1 year ago

consider the titration of hclo4 with koh. what is the ph after 17.0 ml of 0.15 m koh has been added to 15 ml of 0.20 m hclo4?

Chemistry

1 answer:

bezimeni [28]1 year ago

7 0

The ph after 17.0 ml of 0.15 m Koh has been added to 15 ml of 0.20 m hclo4 is 3.347.

Titration is a commonplace laboratory technique of quantitative chemical analysis to determine the attention of an identified analyte. A reagent, termed the titrant or titrator, is ready as a trendy answer of recognized awareness and extent.

Calculation:-

Normality of acid Normality of base

= nMV nMV

= 1 × 0. 15 × 0.017 1 × 0. 20 ×0.015 L

= 2.55 × 10⁻³ = 3 × 10⁻³

The overall base will be high

net concentration = 3× 10⁻³ - 2.55 × 10⁻³

= 0.45 × 10⁻³

= 4.5× 10⁻⁴

pH = -log[4.5 × 10⁻⁴]

= 4 - log4.4

= 3.347

A titration is defined as 'the manner of determining the amount of a substance A by using adding measured increments of substance B, the titrant, with which it reacts till precise chemical equivalence is completed the equivalence factor.

Learn more about titration here:-brainly.com/question/186765

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According to the following reaction, how many grams of potassium sulfate will be formed upon the complete reaction of 23.8 grams

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Answer: The mass of potassium sulfate that can be produced is 73.88 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For KOH:

Given mass of KOH = 23.8 g

Molar mass of KOH = 56.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of KOH}=\frac{23.8g}{56.1g/mol}=0.424mol$

The chemical equation for the reaction of KOH and potassium hydrogen sulfate follows:

$KHSO_4+KOH\rightarrow K_2SO_4+H_2O$

As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.

KOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of KOH produces 1 mole of potassium sulfate

So, 0.424 moles of KOH will produce = $\frac{1}{1}\times 0.424=0.424moles$ of potassium sulfate

Now, calculating the mass of potassium sulfate from equation 1, we get:

Molar mass of potassium sulfate = 174.26 g/mol

Moles of potassium sulfate = 0.424 moles

Putting values in equation 1, we get:

$0.424mol=\frac{\text{Mass of potassium sulfate}}{174.26g/mol}\\\\\text{Mass of potassium sulfate}=(0.424mol\times 174.26g/mol)=73.88g$

Hence, the mass of potassium sulfate that can be produced is 73.88 grams

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