It depends on what compound is dissolved in what solvent.
Presuming a strong monoprotic acid (eg. HCl) in water the pH is equal to -log[H+] = -log(6.8*10^-3) = 3-log6.8 = (3-0.8) = 2.2
Presuming a strong monoprotic base (eg. NaOH) in water the pH is equal to 14 - pOH = 14 - (-log[OH-]) = 14+log(OH-) = 14 + log(6.8*10^-3) = 14 + (-3 + log6.8) = 14 + (-3+0.8) = 14 - 2.2 = 11.8
Fot pH calculation of weak acids and bases you'll need more complex formula and approximations.
Remembwer pH is = the negative logarithm to base ten, of the hydrogen ion concentration . So with a concentration of 0.001 M The hydrogen ion concentration is 0.001 = 10^(-3) ph = -log(10)[H^+] pH = -log(10)10^-3 pH = -(-3) log(10)10 ( Remember log(10)10 = 1 ) pH = -(-3)(1) = --3 = 3 pH = 3
pH = -log[H+] pH = -log[1.6 × 10-3] pH = 2.8
- log(1 X 10 -3 M H +) = 3 pH =====
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
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
- log(2.0 X 10 -3 M) = 2.7 pH ======
Remembwer pH is = the negative logarithm to base ten, of the hydrogen ion concentration . So with a concentration of 0.001 M The hydrogen ion concentration is 0.001 = 10^(-3) ph = -log(10)[H^+] pH = -log(10)10^-3 pH = -(-3) log(10)10 ( Remember log(10)10 = 1 ) pH = -(-3)(1) = --3 = 3 pH = 3
pH = -log[H+] pH = -log[1.6 × 10-3] pH = 2.8
- log(1 X 10 -3 M H +) = 3 pH =====
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
-log(2.3 X 10^-3 M) = 2.6 14 - 2.6 = 11.4 pH ( OH - )
1/103 = 0.001 M ========( pH 3 ) 1/105 = 0.00001 M ============( pH 5 ) As you see, a pH of 3 has a 100 times concentration of 5 pH ( 10 * 10 devalued ) This is the scale; logarithmic.
pH = -log(0.007) = 2.16
3
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
By definition pH = - log10 [H+], where [H+] = concentration H+ of in mol/LSo at [H+] = 7.0*10-3 the answer will be pH = - log (7.0*10-3) = (2.1549) = 2.15
- log(1 X 10^-5 M) = 5 14 - 5 = 9 pH ----------