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Yuliya22 [10]
1 year ago
10

A 0. 22 m aqueous solution of the weak acid ha at 25. 0 °C has a ph of 4. 15. The value of ka for Ha is?

Chemistry
1 answer:
-Dominant- [34]1 year ago
3 0

If 0. 22 m aqueous solution of the weak acid hA at 25. 0 °C has a ph of 4. 15, the value of ka for HA is 2.27 × 10^(-8).

<h3>What is an aqueous solution?</h3>

The aqueous solution is simply defined as something which has been dissolved in water. The aqueous symbol is denoted as (aq).

Given,

pH = 4.15

Concentration of HA = 0.22 M

Firstly, we will calculate the concentration of H+ ion

pH=-log [H+]

4.15 =-log [H+]

= 7.08x 10^{-5}

Secondly we will calculate the value of Ka

for HA is

The equilibrium reaction will be:

HA----- (H+) + (A-)

Given,

Concentration of H+

= Concentration of A-

= 7.08 x 10^{-5} M

The expression of Ka for HA will be,

Ka = [A-] [H+]/[ HA]

= 2.27 × 10^(-8)

Thus we concluded that the value of ka for HA for weak acid HA is 2.27 × 10^(-8).

learn more about aqueous solution or Ka:

brainly.com/question/12481995

#SPJ4

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Luda [366]

The maximum mass of anhydrous zinc chloride that could be obtained from the products of the reaction is 8.18 g

<h3>Stoichiometry </h3>

From the question, we are to determine the maximum mass of anhydrous zinc chloride that could be obtained

From the given balanced chemical equation,

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Learn more on Stoichiometry here: brainly.com/question/11910892

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