The solution for the question above is:
C = 0.270
<span>V = 0.0275L </span>
<span>n = ? </span>
<span>Use the molar formula which is: C = n/V </span>
<span>Re-arrange it to: n = CV </span>
<span>n = (0.270)*(0.0275) </span>
<span>n = 0.007425 mols </span>
<span>(more precise) n = 7.425 x 10^-3 mols
</span>
7.425 x 10^-3 mols is the answer.
The answer you are looking for is True
Answer:
A) pH of Buffer solution = 4.59
B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original buffer solution = 4.65
Explanation:
This is the Henderson-Hasselbalch Equation:
![pH = pKa + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
to calculate the pH of the following Buffer solutions.
Answer:
friend me on here and imma send you the link Explanation: