3.
since the [H+]=0.001 M
then
pH= -log[H+]
-log(0.001)=3
pH=3.
Do you mean 0.001 M H3O ?
-log(0.001 M H3O)
= 3 pH
- log90.00000001 M)
= 8 pH
pH = -log[H3O+] = -log(7.9*10^-11) = 10.1
By definition: pH = -log[H3O+]So pH = -log(7.4*10-9) = 8.13
- log(0.048 M NaOH) = 1.3 pH --------------need Molarity? 1/10 1.3 = 0.050 M H3O ---------------------
- log(1 X 10 -12 M) = 12 pH -----------------------very little room for H3O
pH = 4. The reason is that pH = -log [H+] or -log [H3O+] = -log 10^-4 = 4.
pH = -log[H3O+] = -log(7.9*10^-11) = 10.1
By definition: pH = -log[H3O+]So pH = -log(7.4*10-9) = 8.13
- log(0.048 M NaOH) = 1.3 pH --------------need Molarity? 1/10 1.3 = 0.050 M H3O ---------------------
1.39
- log(1 X 10 -12 M) = 12 pH -----------------------very little room for H3O
1/10 -0.120= 1.32 m h3o+==========
pH = 4. The reason is that pH = -log [H+] or -log [H3O+] = -log 10^-4 = 4.
2 x 10-14 M pH = 13.7
The H3O+ concentration in a solution with pH 3.22 = 1x10^-3.22 M or 6.03x10^-4 M.If a solution is 100 times less acidic, then the H3O+ concentration will be 6.03x10^-6 M.Put another way, 100 times less acidic will have a pH of 5.22 and H3O+ = 1x10^-5.22 = 6.03x10^-6M
If the PH of lemon juice at 298 k is found to be 2.32, the concentration of H3O plus ions in the solution would be 0.5 M.
The answer is 4 - .0001 equals 1X10 -4 which equals a pH of 4
5.0 x 10-3 pH = - log [H3O+] [H3O+] = 1 x 10^-pH pH = 2.3 [H3O+] = 1 x 10^(-2.3) = 5 x 10^(-3) M