6B.3

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Ami_Pant_4G
Posts: 106
Joined: Sat Aug 24, 2019 12:17 am

6B.3

Postby Ami_Pant_4G » Sat Nov 30, 2019 11:01 am

A careless laboratory technician wants to prepare 200.0 mL of a 0.025 m HCl(aq) solution but uses a volumetric flask of volume 250.0 mL by mistake. (a) What would the pH of the desired solution have been? (b) What will be the actual pH of the solution as prepared?

Can someone please explain how to solve part a? thanks in advance.

Nawal Dandachi 1G
Posts: 102
Joined: Sat Sep 28, 2019 12:16 am

Re: 6B.3

Postby Nawal Dandachi 1G » Sat Nov 30, 2019 2:45 pm

Since HCl is a strong acid, then we know that it will 100% dissociate, which tells us that the moles of HCl must be equal to the moles of hydronium ions. So, we can use the volume and molarity given to calculate the moles of hydronium ions present. After calculating the number of moles, divide it by the volume to find the H+ concentration. After that, take the negative log of the concentration to find the pH.

JinwooLee_1F
Posts: 47
Joined: Sat Aug 17, 2019 12:15 am

Re: 6B.3

Postby JinwooLee_1F » Sat Nov 30, 2019 2:55 pm

For problems like these, just know that for strong acids, it will dissociate completely. Therefore, you just have to calculate the pH on the premise that all of the HCl has dissociated.


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