a. What is the theoretical [H+] of 0.00030 M HCl? Answer: 0.00030
b. The above 0.00030 M HCl solution was prepared in the lab by dilution. The pH of the solution was measured as 4.23. Calculate the [H+] from the measured pH.
Answer: pH = 4.23
= 10^ (-4.23)
= 5.88 x 10^-5
c. Calculate the percent error in the measured [H+].
Answer:
(0.00030-(5.88x10^-5))
---------------------- x 100
(0.00030)
= 80%
Am I on the right track?
I agree with all three answers.
Well, you're definitely on the right track, but your calculations are a bit mixed up. Let me guide you through it and add a touch of humor along the way!
a. The theoretical [H+] of 0.00030 M HCl is indeed 0.00030. That's like saying the theoretical number of clowns in a circus is zero if they're all out on vacation!
b. Now, let's calculate the [H+] from the measured pH. If you have a pH of 4.23, you need to remember that pH is a sneaky little devil and represents the negative logarithm of the [H+]. So, to find the [H+], you can do some logarithmic acrobatics:
[H+] = 10^(-pH) = 10^(-4.23) = 5.88 x 10^(-5). It's like finding a hidden treasure in a clown's pocket!
c. To calculate the percent error, you'll need to compare the measured [H+] (5.88 x 10^(-5)) with the theoretical [H+] (0.00030). So the formula is:
Percent Error = (|Theoretical - Measured| / Theoretical) x 100.
Substituting in the values, it becomes:
Percent Error = (|0.00030 - 5.88 x 10^(-5)| / 0.00030) x 100.
After calculations, you'll get a percent error of 80%. That's like juggling flaming bowling pins and only dropping 80% of them! Not too shabby.
Keep up the good work, and remember that chemistry can be fun, just like a clown's antics!
Yes, you are on the right track. The steps you have followed are correct.
In part a, the [H+] of the HCl solution is given as 0.00030 M, so the answer is indeed 0.00030.
In part b, you correctly calculated the [H+] from the measured pH of 4.23 using the formula 10^(-pH). The calculated [H+] is 5.88 x 10^-5.
In part c, you calculated the percent error in the measured [H+]. The formula you used is also correct:
((0.00030 - (5.88 x 10^-5)) / 0.00030) x 100 = 80%
So overall, your steps and calculations are correct.
Yes, you are on the right track! Let's go through each part to confirm your answers.
a. The theoretical [H+] of 0.00030 M HCl is indeed 0.00030. This is because HCl is a strong acid, and when it dissociates in water, it forms equal amounts of H+ and Cl- ions. So, the concentration of H+ in the solution is equal to the concentration of HCl.
b. To calculate the [H+] from the measured pH of 4.23, you need to use the inverse operation of taking the logarithm. This is done by raising 10 to the power of the negative pH value.
pH = 4.23
[H+] = 10^(-pH)
[H+] = 10^(-4.23)
[H+] = 5.88 x 10^(-5)
Your calculation of [H+] as 5.88 x 10^(-5) is correct!
c. To calculate the percent error in the measured [H+], you need to compare the measured value ([H+] from part b) to the theoretical value ([H+] from part a).
Percent error = (|measured value - theoretical value| / theoretical value) x 100
Percent error = (|5.88 x 10^(-5) - 0.00030| / 0.00030) x 100
Percent error = (0.00030 - 5.88 x 10^(-5)) / 0.00030 x 100
Percent error = 0.00024 / 0.00030 x 100
Percent error = 0.8 x 100 = 80%
Your calculation of the percent error as 80% is correct!
Great job on your calculations!