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
To find the pH of the solution created by combining NaOH and HCl, you first need to determine the concentration of the resulting solution. Here is a step-by-step explanation of how to calculate it:
1. Start by calculating the moles of NaOH and HCl:
- NaOH: 11.10 mL x 0.10 M = 0.00111 moles
- HCl: 8.00 mL x 0.10 M = 0.0008 moles
2. Next, subtract the moles of HCl from the moles of NaOH to find the moles of the excess base:
- Excess base (NaOH) = 0.00111 moles - 0.0008 moles = 0.00031 moles
3. To find the total volume of the solution, add the volumes of NaOH and HCl:
- Total volume = 11.10 mL + 8.00 mL = 19.10 mL = 0.0191 L
4. Now, divide the moles of the excess base by the total volume to find the concentration of the resulting solution:
- Concentration = 0.00031 moles / 0.0191 L = 0.0162 M
5. Finally, calculate the pH of the solution using the concentration:
- pH = -log(0.0162) = 1.79
Therefore, the pH of the solution created by combining 11.10 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HCl is approximately 1.79.