pH 1 to 2.
Sulfuric acid is a strong diprotic acid: [H+] = 2* 0.15 = 0.30 mole/L H+
pH = -log 0.30 = 0.52
pH 2.
To calculate the pH, first what you do is find pOH by using the equation pOH= -log[OH] --> pOH= -log[.15M] = .8239. Next, we know that pH+pOH=14. So to find the pH we subtract .8239 which we found earlier from 14--> 14-.8239 = 13.176. So, pH of .15M NaOH= 13.18 (rounded)
0.30102999566398
-log(10^-4 M NaOH) = 4 14 - 4 = 10 pH NaOH -----------------------
1 millimolar = 0.001 M NaOH ( a base, remember ) - log(0.001 M NaOH) = 3 14 - 3 = 11 pH ----------
13
base
A 0.01 M solution of NaOH has a pH =13
13.51
Supposed the acid and base are both strong: pH of the acid is 0.0 and the pH of the base (hydroxide) is 14.0
- log(0.048 M NaOH) = 1.3 pH --------------need Molarity? 1/10 1.3 = 0.050 M H3O ---------------------
na+oh +25=53 ph+poh=2.3
No, sodium hydroxide (NaOH) does not have a pH of 7. Sodium hydroxide is a strong base and has a pH greater than 7. The pH of a solution of sodium hydroxide depends on its concentration. A 0.1 M solution of NaOH has a pH of 13.
Any strong base, such as NaOH ( sodium hydroxide ), with a molarity of at least...., 0.1 M NaOH =============for instance
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE