Calculate pOH from pOH = -log(OH^-).
Then pH + pOH = pKw = 14.
you have to explain these better, i dont 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. thats 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
Very good. That's the first step.
Now step 2 is even easier.
pH + pOH = pKw.
pH + 9.21 = 14.
pH = ??
hmm i know its 5 something
pH + pOH = pKw
pH + 9.21 = 14
pH = 14 - 9.21 = ??
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