what is the pH of an aqeuous solution with s hydroxide ion[OH-] concentration of 7.5x10-10 M?
Calculate pOH from pOH = -log(OH^-).
Then pH + pOH = pKw = 14.
you have to explain these better, i don't even understand them
You must tell me what you don't understand. I don't want to work the problem for you. All you need to do is to plug in OH^- and calculate pOH and from that pH. Two algebraic equations that don't get any simpler.
im struggling in this class, and i need as much help as possible.
I have plenty of help and I am pleased to help you as much as I can but I don't do homework. Again, I won't work the problem for you but tell me what you don't understand and we can get started on it.
What do you not understand about
pOH = -log(OH^-)?
i understand your trying to help me..but i don't understanf anything that you have helped me.im a step by step person. that's why i thought this would work
I'm glad you are a step by step person. So am I. I gave two steps.
Step 1. pOH = -log(OH^-).
1a. Punch in the OH^- in the problem into your calculator.
1b. Hit the log key.
1c. The answer jumps onto the screen.
1d. Change the sign.
1e. That's the pOH.
Step 2. pH + pOH = pKw.
2a. You know pOH from step 1.
2b. You know pKw, or at least I assume you do. pKw is 14.
2c. Now you have the second equation which is pH + the number for pOH = 14.
2d. Now you solve for pH.
2e. You have your answer and you've worked the problem.
so is this what it should look like
pOH=-log(7.5x10-10)=9.124
Very good. That's the first step.
Now step 2 is even easier.
pH + pOH = pKw.
pH + 9.21 = 14.
pH = ??