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
For 0.1 m NaOH, pOH = -log[OH-] = -log(0.1) = 1
pH + pOH = 14, So pH = 13
Same way for 0.05 m NaOH, pOH = -log(0.05) = 1.3
pH = 14 - 1.3 = 12.7
pH=0
-log(10^-4 M NaOH) = 4 14 - 4 = 10 pH NaOH -----------------------
3
As NaOH is a strong base I would not be surprised to see a 14 pH at least.
The pH value for this solution is 13.
Sodium hydroxide, NaOH, has a very high pH, so it can be used to raise the pH of some mixture that is too acidic.
You dont - adding NaOH increases pH.
1 millimolar = 0.001 M NaOH ( a base, remember ) - log(0.001 M NaOH) = 3 14 - 3 = 11 pH ----------
-log(10^-4 M NaOH) = 4 14 - 4 = 10 pH NaOH -----------------------
3
As NaOH is a strong base I would not be surprised to see a 14 pH at least.
HCl : makes it acidic. it decreases the pH NaOH : makes it alkaline. it increases the pH
The pH value for this solution is 13.
Sodium hydroxide, NaOH, has a very high pH, so it can be used to raise the pH of some mixture that is too acidic.
A 0.01 M solution of NaOH has a pH =13
Quantity matters. If there is a lot of buffer (in terms of moles) and relatively little NaOH then the buffer will prevent any change in pH. If there is relatively more NaOH than buffer, then of course the pH will rise.
The amount of NaOH needed to raise the pH from 8 to 10 depends heavily on the conditions. The amount of NaOH needed will increase as the volume of the solution increases. Even more importantly, buffers can stabilize the pH significantly. If buffers are presently, the pH change will be much more gradual, and more NaOH will be required.
13