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Verdich [7]
3 years ago
10

Whch is more acidic, a solution with a pH of 2 or 5

Chemistry
2 answers:
Molodets [167]3 years ago
3 0

Answer:

A solution with a pH of 2

Explanation:

The lower a pH is in a liquid, the more acidic it is. 7 is the neutral pH for a liquid. We usually drink around that pH for water. A pH higher than 7 expresses that the liquid is a base.

Hoochie [10]3 years ago
3 0

Answer:

A solution with a pH of 2 is more acidic.

Explanation:

The solution with a pH of 2 is more acidic because on the pH scale as you get closer to zero, the more acidic the solution becomes.

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Crush the limestone... it would give more area for the acid to react
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Which of the following acids should be used to prepare a buffer with a pH of 4.5?A. HOC6H4OCOOH, Ka = 1.0 x 10^-3B. C6H4(COOH)2,
ioda

Answer:

C. CH3COOH, Ka = 1.8 E-5

Explanation:

analyzing the pKa of the given acids:

∴ pKa = - Log Ka

A. pKa = - Log (1.0 E-3 ) = 3

B. pKa = - Log (2.9 E-4) = 3.54

C. pKa = - Log (1.8 E-5) = 4.745

D. pKa = - Log (4.0 E-6) = 5.397

E. pKa = - Log (2.3 E-9) = 8.638

We choose the (C) acid since its pKa close to the expected pH.

⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):

  • pH = pKa + Log ([CH3COO-]/[CH3COOH])

∴ pH = 4.5

∴ pKa = 4.745

⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])

⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])

⇒ 0.5692 = [CH3COO-]/[CH3COOH]

∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]

⇒ 1.8 E-5 = [H3O+](0.5692)

⇒ [H3O+] = 3.1623 E-5 M

⇒ pH = - Log ( 3.1623 E-5 ) = 4.5

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3 years ago
Which surface ocean currents has the warmest water
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Does the universe have air in it between the galxies in it
fenix001 [56]

Answer:

No it does not space does not have oxygen and air is caused by oxygen

hey but this was a really interesting question to ask tho

Explanation:

4 0
3 years ago
Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of chromium(III) bromide and pota
expeople1 [14]

Explanation:

An ionic equation will be the one in which all the participating species will be present as ions.

The given reaction will be as follows.

     CrBr_{3} + K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + KBr

Balancing this equation by multiplying CrBr_{3} by 2 and K_{2}CO_{3} by 3 on reactant side. Whereas multiply KBr by 6 on product side.

      2CrBr_{3} + 3K_{2}CO_{3} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6KBr

Hence, the net ionic equation will be as follows.

       2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) + 6K^{+} + 6Br^{-} \rightarrow Cr_{2}(CO_{3})_{3}(s) + 6K^{+} + 6Br^{-}

As both K^{+} and Br^{-} are spectator ions. Hence, the net ionic equation will be as follows.

     2Cr^{3+}(aq) + 3CO^{2-}_{3}(aq) \rightarrow Cr_{2}(CO_{3})_{3}(s)  

4 0
3 years ago
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