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Calculating pH?

400mL of 0.2M aqueous solution of NaOH was mixed with 300mL of 0.25M HBr solution. Is the resulting solution acidic or basic? Calculate molar concentrations of all ions in this solution. Ignore self-dissociation of water.

So first I did,

HBr + NaOH --> NaBr + H20

0.075 moles of HBr reacts with 0.08 moles of NaOH.
Since NaOH is the limiting reagent, following 1:1 ratio, 0.075 moles of NaBr and H2O are produced.

Water doesn't affect the pH, HBr is gone, so Im left with either NaOH or NaBr to determine the pH value. Which one do I choose?
HBr is the limiting agent because there’s less moles of it. So you use what’s left of the NaOH to calculate the pH
Calculate the final concntration of OH-.
Then use the ionic product of water to find the H+ concentration; and from that the pH.

I have written in more detail how to use the ionic product of water here http://alevelchemistrytuition.co.uk/ionic-product-of-water/
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LLearnt
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Original post by TutorsChemistry
Calculate the final concntration of OH-.
Then use the ionic product of water to find the H+ concentration; and from that the pH.

I have written in more detail how to use the ionic product of water here http://alevelchemistrytuition.co.uk/ionic-product-of-water/


Regarding this question, if I did the following would I be correct:

0.075x1000/700= Concentration

-log[concentration] = ph

And as its the first time im going over a limiting reagent, is it the one that is not in excess so has reacted fully?
Thanks

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