## Error in 6B.5 e)?

Daniel_Frees_1L
Posts: 40
Joined: Fri Sep 28, 2018 12:18 am

### Error in 6B.5 e)?

I've gotten the same answer for 6B.5e) over and over and I'm convinced im doing the problem right. mg/L --> dimensional analysis to mol/L. OH- Molarity is the same because strong base. -log(M OH-) = -log(0.009714) = 2.012 = pOH. pH = 11.998. Does the answer key have a typo, or am I missing something?

Chem_Mod
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### Re: Error in 6B.5 e)?

Can you please post the full question so that everyone can have it for reference?

VPatankar_2L
Posts: 84
Joined: Thu Jul 25, 2019 12:17 am
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### Re: Error in 6B.5 e)?

Make sure you have the correct decimal places for your molarity. 13.6 mg --> 0.0136 g
moles NaOH = (0.0136 g)/(40 g/mol) = 3.4 * 10^-4 mol
Molarity = (3.4 * 10^-4 mol)/(0.350 L) = 9.7143 *10^-4 mol/L
pOH = -log [9.7143 *10^-4 mol/L] = 3.01

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