Asked by danny
What is the pH of 500 mL of water that had 0.00150 grams of NaOH added to it? The units are pH units with an error interval of 0.1 units.
0.00150 divided by 40NaOH 3.75e-5 divided that by 500= 7.5e-8 -log7.5e-8 =7.12 im lost on the last step
Answers
Answered by
DrBob222
Not quite. Thanks for showing your work.
0.00150/40 = mols KOH and that's right. But that's divided by L (0.500) and not 500. That give you about 7.4E-5 and that is the (KOH) in the solution. Take the -log of that to give pOH and subtract from 14 to obtain the pH.
0.00150/40 = mols KOH and that's right. But that's divided by L (0.500) and not 500. That give you about 7.4E-5 and that is the (KOH) in the solution. Take the -log of that to give pOH and subtract from 14 to obtain the pH.
Answered by
danny
is the answer 9.86923
Answered by
DrBob222
No but I would go with 9.87
I don't know what you're supposed to do with the error unit of 0.1.
I don't know what you're supposed to do with the error unit of 0.1.
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