Assuming: pH (potentia hydrogenii) = - log10(H+ ionic concentration in mol/Liter). For a buffer solution containing acetic acid and sodium acetate, according to Henderson-Hasselbalch equation, the solution will have pH = 4.7 + log10(concentration of acetic acid / concentration of sodium acetate). If some sodium hydroxide (NaOH) is added that much (in mol for mol) acetic acid will be converted to sodium acetate. Your function should output final pH for a small input of NaOH in mol.
For example, when NaOH added is 0 mol to a solution containing 0.1 mol acetic acid and 0.1 mol sodium acetate then final pH is 4.7.
Solution Stats
Solution Comments
Show commentsProblem Recent Solvers55
Suggested Problems
-
Remove any row in which a NaN appears
8790 Solvers
-
Maximum running product for a string of numbers
2258 Solvers
-
2348 Solvers
-
07 - Common functions and indexing 6
468 Solvers
-
529 Solvers
More from this Author99
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!