chemistry steps check
posted by Amy on .
What is the pH of the solution created by combining 11.10 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
This is exactly like what Dr Bob has shown me on the first problem i put here and yet i get this simple one wrong...
Here are my steps:
NaOH: 0.0111L x.1 M= .00111 moles
HCl: 0.008L x .1M=.0008 moles
.0011 .0008= .0003
.0003/ .0191= .0157
pH= log (.0157)= 1.80

See my comments in bold below
What is the pH of the solution created by combining 11.10 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
This is exactly like what Dr Bob has shown me on the first problem i put here and yet i get this simple one wrong...
Here are my steps:
NaOH: 0.0111L x.1 M= .00111 moles
HCl: 0.008L x .1M=.0008 moles
.0011 .0008= .0003 I believe you droped a 1 here.
0.00111  0.0008 = 0.00031
.0003/ .0191= .0157 So this becomes 0.01623 which gives a number close to 1.80 but not quite?? Confirm.
pH= log (.0157)= 1.80 
k so i put 1.79 and 1.18 and the system still says its wrong.

Well, I did NOT make an error in my calculations. But I goofed big time. Notice that the NaOH is the excess, not HCl. (There is more NaOH to begin than there is HCl.) Therefore, the 0.00031 is the mol NaOH left unreacted. That means the 1.79 you calculated is the pOH. Just subtract that from 14.0. Remember, pH + pOH = 14. I get 12.21 but check me out on that. And check my arithmetic. Sorry I didn't read the problem more carefully.

alright i got this one.. thank you