1. Calculate 1) the pH of a solution which is 0.15 M in HCl and 0.05 M in NaOH and 2) the pH of a solution which is 0.05 M in HCl and 0.15 M in NaOH.
This solution can be considered equivalent to a solution 0.1 M (0.15 - 0.05) in HCl thus
pH = -Log (0.1) = 1
In the same way a solution 0.05 M in HCl and 0.15 M in NaOH, can be considered equivalent to a solution 0.1 M in NaOH thus
pOH = -Log (0.1) = 1pH = 14 - 1 = 13
2. Calculate the pH of a solution 1M in HCl and 1M in NaOH.
In this case the only source of H3O+ is water dissociation thus pH = 7 as in pure water.
Note: The solution above is equivalent to a solution 1M in NaCl, i.e., the salt formed by the strong acid HCl and the strong base NaOH.
3. Calculate the pH of a solution obtained by mixing 500 ml of 0.1 M HCl and 500 ml of 0.05 M ammonia (NH3).
500 ml of 0.1 M HCl contains 0.05 moles of the acid, after mixing the volume is 1 liter thus HCl concentration becomes 0.05 M. Analogously NH3 concentration becomes 0.025 M. The reaction of neutralization can be written as:
NH3+ H+ --> NH4+
This reaction will produce 0.05 - 0.025 = 0.025 moles/liter of NH4+ leaving 0.025 moles/liter of H+. Therefore:
pH = - log 0.025 = 1.6
4. Calculate the pH of one liter of solution prepared by mixing 1.3 moles of acetic acid (pKa 4.75) and 0.6 moles of NaOH.
[CH3COO-] = 0.6 M (the concentration of the base added)
[CH3COOH] = 1.3 - 0.6 = 0.7 M
pH = pKa - Log [CH3COOH]/[CH3COO-] = 4.75 - Log (0.7) + Log (0.6) = 4.68
5. Calculate the pH of one liter of solution containing 0.6 moles of sodium acetate and 0.7 moles of acetic acid.
This problem is perfectly equivalent to the one above. In fact, the dissociation of sodium acetate (strong electrolytes) produce 0.6 moles of acetate. Thus, we will have a solution containing 0.6 moles of acetate and 0.7 moles of acetic acid. In this case the equation used above assumes the following form:
pH = pKa - Log Ca/Cs
where Ca is the concentration of the acid and Cs the concentration of the salt. Thus
pH = 4.75 - Log (0.7) + Log (0.6) = 4.68