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Chemistry 13 Online
OpenStudy (anonymous):

Will fan + give medal. How would you solve for pH of 2.5 x 10-2 M HCl? What would you need to do if you were given the concentration of hydroxide ion rather than hydronium?

OpenStudy (jfraser):

what's the relationship between the \(H^{+1}\) concentration and pH?

OpenStudy (anonymous):

I have no idea, I just need the answer for a study guide.

OpenStudy (jfraser):

if I give you the answer for this study guide, and the next time you need to use this study guide, but the numbers are different, you'll have no idea what to do, so you really need to know how to solve the problem, not just what this particular answer happens to be. ok?

OpenStudy (anonymous):

Not really because this is my last test and my only type of this question.

OpenStudy (jfraser):

well if you want to learn, i'll show you how. If someone wants to just give you the answer, I'll let them do that. Good luck, though

OpenStudy (anonymous):

Yes, I agree. You need to know how to do this. Being given the right answer will not help you further on.

OpenStudy (anonymous):

Okay, so what do I do then?

OpenStudy (anonymous):

@JFraser

OpenStudy (anonymous):

First, you need to know that hydrochloric acid is a strong acid. It disassociates quantitatively. It produces one hydronium ion for each molecule of HCl that dissociates. So, now that you know the concentration of the hydronium ion in solution, find pH using:\[\color{green}{\text{pH} = -\log [\text{H}_3\text{O}^{+}]}\]

OpenStudy (anonymous):

pH= -log(2.5 x 10^-2) pH=-(-1.60) pH=1.60 Is that right?

OpenStudy (anonymous):

And for the second part: When you are given the concentration of hydroxide ions in a basic solution, you can use that value to find the concentration of the hydronium ions present using the following relationship: [H3O+] × [OH-] = 1.0 × 10-14 M The value 1.0 × 10-14 is called the ionization constant of water. The relationship above can be used to compare the hydronium and hydroxide ion concentrations in any aqueous solution at 25°C.

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