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Varvara68 [4.7K]
3 years ago
8

A buffer with a pH of 3.98 contains 0.23 M of sodium benzoate and 0.38 M of benzoic acid. What is the concentration of [H3O+] in

the solution after the addition of 0.058 mol HCl to a final volume of 1.3 L? Assume that any contribution of HCl to the volume is negligible g
Chemistry
1 answer:
Vika [28.1K]3 years ago
3 0

Answer:

New pH = 3.84

Explanation:

First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.

Let's determine the moles of each compound:

0.23 M . 1.3L = 0.299 moles of NaBz

0.38 M . 1.3L = 0.494 moles of HBz

We add 0.058 of HCl, which is the same as 0.058 moles of H⁻

HCl →  H⁺  +  Cl⁻

As we add the moles of protons, these are going to react to the Bz⁻

In the buffer system we have these dissociations:

NaBz  →  Na⁺  +  Bz⁻

HBz →  H⁺  + Bz⁻

So, as we add protons, we have a new equilibrium:

          Bz⁻      +      H⁺       ⇄   HBz    

In    0.299         0.058          0.494

Eq   0.241               -              0.552

Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation

pH = pKa + log (Bz⁻/HBz)

pH = 4.20 + log (0.241 / 0.552) → 3.84

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Consider the following reaction:
slava [35]

Answer:

Option D is correct =  58 g

Explanation:

Data Given:

mass of LiOH = 120 g

Mass of Li3N= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

                  Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)

                    1 mol                                                      3 mol

Convert moles to mass

Molar mass of LiOH = 24 g/mol

Molar mass of  Li₃N = 35 g/mol

So,

                   Li₃N (s)        +     3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)

                1 mol (35 g/mol)                                                   3 mol (24 g/mol)

                   35 g                                                                         72 g

So if we look at the reaction 35 g of Li₃N react with water and produces 72 g of LiOH , then how many g of Li₃N will be react to Produce by 120 g of  LiOH

For this apply unity formula

                        35 g of  Li₃N ≅ 72 g of LiOH

                        X of  Li₃N ≅ 120 g of LiOH

By Doing cross multiplication

                  Mass of Li₃N = 35 g x 120 g / 72 g

                   mass of Li₃N =  58 g

120 g of LiOH will produce from 58 g of Li₃N

So,

Option D is correct =  58 g

5 0
3 years ago
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