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he hydrogen ion concentration of a \[{10^{ - 8}}M\] HCl aqueous solution at $298K$ is:
Given that: ${K_w} = {10^{ - 14}}$
A.$1.0 \times {10^{ - 6}}M$
B.$1.0525 \times {10^{ - 7}}$
C.$9.525 \times {10^{ - 8}}$
D.$1.0 \times {10^{ - 8}}M$

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Last updated date: 23rd Feb 2024
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IVSAT 2024
Answer
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Hint: We need to calculate the Hydrogen ion concentration in the HCl aqueous solution at a given temperature. Now we will know the ${K_w}$ of water. The water's behavior is neutral so we will know the concentration of ${H^ + }$ ion. Now we will know the total molarity of ${H^ + }$ ion in HCL. Now add the concentration of ${H^ + }$ in HCl and in water. We will get the final concentration.

Complete Step by step answer: Step1. Write what is given in the question:
Molarity of HCL ion: ${10^{ - 8}}M$
Given the concentration of ${H^ + }$ in water is ${10^{ - 14}}$ .
Temperature given is $298K$ . This is normal room temperature.
Step2. ${K_w}$ of ${H_2}O$ = ${10^{ - 14}}$
But it contains both the hydrogen ion and hydroxide ion. So we can say that the concentration of hydrogen ions will be ${10^{ - 7}}$ due to its neutral behavior. The amount of hydroxide ion is the same as the hydrogen ion.
${H^ + }$ ion concentration in water will be ${10^{ - 7}}$
Step3. In the aqueous solution of ${10^{ - 8}}$ $M$ of HCl,
Total ${H^ + }$ will be ${H^ + }$of HCl + ${H^ + }$ of water.
$ \Rightarrow {10^{ - 8}} + {10^{ - 7}}$
$ \Rightarrow 11 \times {10^{ - 8}}M$
$ \approx 1.10 \times {10^{ - 7}}M$

Hence the option B is the correct answer.

Note: The concentration of hydrogen ion is generally considered as the pH, which is calculated as the logarithm of the reciprocal of the hydrogen ion concentration in gram moles per liter. The water is the neutral so the concentration of hydrogen ion and hydroxyl ion is equal. If the pH is less than 7 than it will be basic and if the pH is greater than 7 than it will be acidic.
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