Calculate the pH of each of the following solutions:
a) .35 M hydrochloric acid, HCl
b) .35 M acetic acid, HC2H3O2
c) .35 M sodium hydroxide, NaOH
Can someone please explain to me how to calculate the pH in a way that is easy to understand? I am confused. Could you please do one of the above questions as an example? Thank you so much!
a) 0.35M HCl (monoprotic strong acid) produces a 0.35M [H+] solution.
pH = -log(0.35)
c) 0.35M NaOH (strong base) acid produces a 0.35M [OH-] solution.
pOH = -log(0.35)
pH = 14-pOH
b) 0.35M HC2H3O2 (WEAK ACID) does not dissociate completely. Its Ka = 1.8x10^-5
[H+] = sqrt[(Ka)(0.35)]
pH = -log[H+]
This has a very good explaination. Go through his examples, especially at the end.
On c, find the OH concentration, then the pOH, then you know pH+pOH=14 and you can solve for pH from that.
True
Of course! I can certainly help you understand how to calculate the pH of a solution in a simple way. Let's take the first example and calculate the pH of a 0.35 M hydrochloric acid solution, HCl.
To calculate the pH of an acid solution, you need to consider the concentration of hydrogen ions, H⁺, in the solution. In the case of hydrochloric acid (HCl), it is a strong acid that dissociates completely in water, meaning it releases all of its hydrogen ions into the solution.
Step 1: Identify the concentration of the hydrogen ions (H⁺) in the solution. In this case, the concentration of H⁺ is the same as the concentration of HCl, which is 0.35 M.
Step 2: Take the negative logarithm (base 10) of the hydrogen ion concentration. Mathematically, this is represented as pH = -log[H⁺].
So, for the given HCl solution with a concentration of 0.35 M, we can calculate the pH as follows:
pH = -log[0.35]
To complete this calculation, you would need to use a scientific calculator or look up the logarithm table. The result of this calculation is approximately pH = 0.455.
Therefore, the pH of a 0.35 M hydrochloric acid solution (HCl) is approximately 0.455.
You can follow the same steps to calculate the pH for the other examples you've mentioned, namely, a 0.35 M acetic acid (HC2H3O2) solution and a 0.35 M sodium hydroxide (NaOH) solution.