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omeli [17]
3 years ago
13

what happens to change the solubility when you added 6.0 M HCl to the second test tube with benzoic acid

Chemistry
1 answer:
shusha [124]3 years ago
8 0

Answer:

It becomes insoluble again.

Explanation:

The chemical equation showing the reaction between benzoic acid, and sodium hydroxide,NaOH is given below;

C6H5COOH + NaOH ------------------------> C6H5COO^- Na^+ + H2O.

The Benzoic acid, C6H5COOH is insoluble but when it reacted with 1M Sodium hydroxide, NaOH it changes to sodium benzoate, C6H5COO^- Na^+ which is more soluble than the benzoic acid.

C6H5COO^- Na^+ + HCl -------------------> C6H5COOH + NaCl.

When the sodium benzoate, C6H5COO^- Na^+ reacts with 6M HCl, it is converted back to benzoic acid which is insoluble. Hence, precipitation is observed.

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Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.
Softa [21]

Explanation:

PbBr_{2} will dissociate into ions as follows.

         PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)

Hence, K_{sp} for this reaction will be as follows.

                   K_{sp} = [Pb^{2+}][Br^{-}]^{2}

We take x as the molar solubility of PbBr_{2} when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

               [Pb^{2+}] = [Pb^{2+}]_{o} + x

               [Br^{-}]^{2} = [Br^{-}]_{o} + 2x

So, equilibrium solubility expression will be as follows.

            K_{sp} = ([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains [Br^{-}]_{o} = 0.10 and there will be no lead ions. So, [Pb^{2+}]_{o} = 0

So, [Br^{-}]_{o} + 2x will approximately equals to [Br^{-}]_{o}.

Hence, K_{sp} = x[Br^{-}]^{2}_{o}

            4.67 \times 10^{-6} = x \times (0.10)^{2}

                        x = 4.67 \times 10^{-4} M

Thus, we can conclude that molar solubility of PbBr_{2} is 4.67 \times 10^{-4} M.

5 0
3 years ago
Read 2 more answers
KOH + HBr - KBr + H2O<br> Which is the acid in this reaction?
Reptile [31]

Answer:

HBr is a strong acid

Explanation:

KBr is a salt which makes a base . also KOH is a base

7 0
3 years ago
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1) What is the formula for calculating weight
MatroZZZ [7]

Weight we calculate using W= force multiplied to displacement

Mass SI unit is Kg

Weight SI unit is Newton (N)

Mass

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3 years ago
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Ne4ueva [31]

Answer:

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4 0
3 years ago
The molar solubility of ag2s is 1.26 × 10-16 m in pure water. calculate the ksp for ag2s.
MAXImum [283]
Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq)<span>.
</span>s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.
6 0
3 years ago
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