- #1
patrickbotros
- 34
- 1
This is an example in my book. I have the answer for it, but I don't know why my answer was wrong. Here's the question:
Lauren pours 40 mL of .1M NaOH into 100 mL of Megan's .1 M HCl. What is the pH?
Here's what I did:
But my teacher's book says pH=1.37.
UPDATE: I was using WolframAlpha.com to calculate the log and apparently it calls log what I would call ln. When I entered log base 10 (.006/.14) everything worked out fine. If somebody knows how, delete this post please.
Lauren pours 40 mL of .1M NaOH into 100 mL of Megan's .1 M HCl. What is the pH?
Here's what I did:
R NaOH+HCl ⇔ NaCl + H2O
I .004 , .01 ⇔ 0 0
C -.004 , -.004 ⇔ +.004 , +.004
E 0 , .006 ⇔ .004 , .004
Then, I made another RICE table for the remaining HCl:I .004 , .01 ⇔ 0 0
C -.004 , -.004 ⇔ +.004 , +.004
E 0 , .006 ⇔ .004 , .004
R HCl + H2 O ⇔ H3O++ Cl-
I .006
C -.006 -.006 +.006 +.006
So then pH=-log(.006/.14)=3.14I .006
C -.006 -.006 +.006 +.006
But my teacher's book says pH=1.37.
UPDATE: I was using WolframAlpha.com to calculate the log and apparently it calls log what I would call ln. When I entered log base 10 (.006/.14) everything worked out fine. If somebody knows how, delete this post please.
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