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1 year ago

Calculate the hydrogen ion concentration in a solution of fruit juice having a pH of 4.25.

Chemistry

1 answer:

Airida [17]1 year ago

3 0

Answer:

<h3>The answer is option B</h3>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ H+ ]

To find the hydrogen ion concentration substitute the pH into the above formula and solve for the [ H+ ]

From the question

pH = 4.25

So we have

4.25 = - log [ H+ ]

Find the antilog of both sides

That's

We have the final answer as

<h2> $[ H+ ] = 5.6 \times {10}^{ - 5} \: M$ </h2>

Hope this helps you

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Art [367]

I am sure the answer is A

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1 year ago

Read 2 more answers

Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A

lesantik [10]

Answer: The chemical equations and equilibrium constant expression for each ionization steps is written below.

Explanation:

The chemical formula of carbonic acid is $H_2CO_3$ . It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

$H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)$

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}$

The chemical equation for the second dissociation of carbonic acid follows:

$HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)$

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}$

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.

6 0

2 years ago

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$