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3 years ago

A 80.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 40.0 mL of KOH at 25 ∘C.

1 answer:

4 0

Number of moles of HBr = 80.0 mL x(1L / 1000 mL) x (0.25 mol / L) = 0.02 mol
Number of moles of KOH = 40.0 mL x (1 L / 1000 mL) x (0.5 mol / L) = 0.02 mol
At this point all the acid is completely neutralized by base so; at the equivalence point pH of solution equal 7 (salt of strong acid and strong base KBr)

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Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,

ludmilkaskok [199]

Answer: The empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

Explanation:

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = $\frac{5.80}{1.44}=4.03\approx 4$

For Oxygen = $\frac{1.44}{1.44}=1$

For Nitrogen = $\frac{1.44}{1.44}=1$

For Chlorine = $\frac{1.44}{1.44}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

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