Ask your own question, for FREE!
Chemistry 9 Online
OpenStudy (anonymous):

Find the pH and pOH of 10^-2 M of HCI

OpenStudy (anonymous):

I really need help with this question, I have an exam tomorrow morning and this is on the review and I have no idea how to solve it.

OpenStudy (abhisar):

Hello @christina2797 ! \(\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\\\color{white}{.}\\\Huge\sf\color{blue}{~~~~Welcome~to~OpenStudy!~\ddot\smile}\\\color{white}{.}\\\\\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\)

OpenStudy (abhisar):

Since HCL is a strong acid, it completely ionizes. Thus 10^-2 M of HCL will give the same amount of H+ ions

OpenStudy (anonymous):

can you help me? @Abhisar

OpenStudy (abhisar):

Now pH = -log(H+)

OpenStudy (anonymous):

Okay so pH= -log(10^-2)= 2

OpenStudy (abhisar):

So the pH of HCL will be \[-\log \left( 10^{-2} \right)\]

OpenStudy (abhisar):

yes u got it

OpenStudy (anonymous):

and then from there I just do 14-2 right? to find the pOH?

OpenStudy (abhisar):

yes pH + pOH = 14

OpenStudy (anonymous):

Thank you so so much!

OpenStudy (abhisar):

\(\Huge\text{Anytime !}\) \(\huge\ddot\smile\)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!