Re: CHEMISTRY HW! IM DESPERATE
galopus galopus <[email protected]>
| Newsgroups | gmane.science.chemistry.the-chemistry-cluster |
|---|---|
| Message-ID | <[email protected]> |
First of all you have to find out the amount of moles of one substance that left
over after the neutralization:
equall amount (moles) of each neutralizes each other: 10 mL fo 0.182 M HNO3
39.10 mL sample of 0.413 M NaOH
Number of moles: 0.182 mol/L x 0. 01 L = 0.00182 mol = 0.002 mol
(acid)
0.039.1 L x 0.413 mol/L = 0.016
mol (base)
Remain substansce= 0.016 mol- 0.002 mol = 0.014 mol of base
total volumen = 39.1 mL + 10 mL = 49.1 mL = 0.049 L
So the molar concentration od OH is= 0.014 mol/0.049 L = 0.286 mol/L
pOH = -log (OH) = -log 0.286 = 0.544
pH + pOO = 14
pH = 14- 0.544 = 13.46
I hpoe i did help. I did it with 3 decimals because i never saw a better way.
Saludos
________________________________
From: bubby2250 <[email protected]>
To: [email protected]
Sent: Thu, November 11, 2010 11:51:35 AM
Subject: [The Chemistry Cluster] CHEMISTRY HW! IM DESPERATE
A 10.00 mL sample of 0.182 M HNO3 is titrated with a 39.10 mL sample of 0.413 M
NaOH at 25 oC. What is the pH of the resulting solution? Assume that the volumes
are additive. (Express your answer to two places behind the decimal.)
[Non-text portions of this message have been removed]
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