Aqueous Solutions¶

Acidity Calculation (pH, pOH)¶

chemlib.chemistry.pH(**kwargs)¶

For any inputted pH, pOH, [H+], or [OH-], finds the corresponding values.

Parameters: kwargs – The value of the chosen input (pH=, pOH=, H=, or OH=)
Return type: dict

What is the pH, pOH and [OH-] given a [H+] of 1.07x10^-6 M?

>>> import chemlib
>>> chemlib.pH(H=1.07e-6)
{'H': 1.07e-06, 'pOH': 8.029, 'pH': 5.971, 'OH': 9.354e-09, 'acidity': 'acidic'}

What is the pH, pOH, and [H+] given a [OH-] of 2.06x10^-3 M?

>>> chemlib.pH(OH=2.06e-3)
{'OH': 0.002, 'H': 5e-12, 'pOH': 2.699, 'pH': 11.301, 'acidity': 'basic'}

What is the pOH, [H+], and [OH-] given a pH of 5.2?

>>> chemlib.pH(pH = 5.2)
{'pH': 5.2, 'OH': 1.585e-09, 'H': 6.309e-06, 'pOH': 8.8, 'acidity': 'acidic'}

Making a Solution¶

class chemlib.chemistry.Solution(self, solute, molarity)¶

Instantiate a chemlib.Solution object with the formula of the solute, and the molarity in mol/L.

param solute (str)

The formula of the solute without using subscripts OR a chemlib.chemistry.Compound object.

param molarity (float)

How many moles of solute per liter of solution

>>> from chemlib import Solution
>>> Solution('AgCl', 2)
<chemlib.chemistry.Solution object at 0x03F46370>
>>> s = Solution('AgCl', 2)
>>> s.molarity
2

classmethod chemlib.chemistry.Solution.by_grams_per_liters(cls, solute, grams, liters)¶

OR you can make a Solution with the solute, and grams per liter.

param str solute

The formula of the solute without using subscripts OR a chemlib.chemistry.Compound object.

param float grams

How many grams of solute

param float liters

How many liters of solution

>>> from chemlib import Solution
>>> s = Solution.by_grams_per_liters("NaCl", 10, 1)
>>> s.molarity
0.1711

Dilutions¶

chemlib.chemistry.Solution.dilute(self, V1=None, M2=None, V2=None, inplace=False) → dict¶

Using formula M1*V1 = M2*V2

param float V1

The starting volume of the solution. [Must be specified]

param float M2

The ending molarity after dilution.

param float V2

The ending volume after dilution

param bool inplace

You can set to true if the old molarity is to be replaced by the new molarity

return

The new volume and the new molarity.

rtype

dict

raises TypeError

if a starting volume is not specified

raises TypeError

if both M2 and V2 are specified

To find the dilution of 2.5 L of 0.25M NaCl to a 0.125M NaCl solution:

>>> from chemlib import Solution
>>> s = Solution("NaCl", 0.25)
>>> s.dilute(V1 = 2.5, M2 = 0.125)
{'Solute': 'Na₁Cl₁', 'Molarity': 0.125, 'Volume': 5.0}