PH of 0.015 M HCL is?

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PH of 0.015 M HCL is?

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Answer 1 · 2018-04-06T04:10:30

I get $1.82$ $1.82$ .

Explanation:

$pH$ $"pH"$ is given by the equation,

$pH = - log [H^{+}]$ $"pH"=-log[H^+]$

$[H^{+}]$ $[H^+]$ is the hydrogen ion concentration in terms of molarity.

Since hydrochloric acid is a strong acid, it dissociates into single $H^{+}$ $H^+$ ions in an aqueous solution, which eventually become $H_{3} O^{+}$ $H_3O^+$ ions due to the presence of water $(H_{2} O)$ $(H_2O)$ , and the $pH$ $"pH"$ will simply be:

$pH = - log (0.015)$ $"pH"=-log(0.015)$

$\approx 1.82$ $~~1.82$

Answer 2 · 2018-04-06T04:11:34

See below.

Explanation:

$H C l$ $HCl$ is a strong acid. That means that it dissociates completely into $H^{+}$ $H^+$ and $C l^{-}$ $Cl^-$ :

${HCl}_{(a q)} \to H^{+} + {Cl}^{-}$ $"HCl"_((aq)) -> "H"^+ + "Cl"^-$

This means that the initial concentration of $H C l$ $HCl$ is equal to the equilibrium concentration of $H^{+}$ $H^+$ .

Therefore, $[HCl] = [H^{+}]$ $["HCl"]=["H"^+]$ , and $[H^{+}] = 0.015 M$ $["H"^+]=0.015" M"$ at equilibrium.

Here is an equation that you will need to be familiar with in order to solve this problem:

$p H = - log [H^{+}]$ $pH=-log["H"^+]$

Now, we take our concentration of $[H^{+}] = 0.015 M$ $["H"^+]=0.015" M"$ and plug it in as the equilibrium concentration of $[H^{+}]$ $["H"^+]$ at equilibrium in the equation, and solve for the $p H$ $pH$ . Here is the setup to obtain the final answer:

$p H = - log [0.015]$ $pH=-log[0.015]$

Plug this into your calculator, and the answer will be $1.82$ $color(red)(1.82$ , with significant figures (note that the digit $1$ $1$ in the answer does not count as a significant figure. If you need help with significant figures and pH, please leave a comment.)

I hope that helps!

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PH of 0.015 M HCL is?

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