Dilution and Mixing

*Problem 1: Dilution of Sulfuric Acid*

40 mL of 2.5 molar H_{2}SO_{4} solution will be diluted to 2 L. What is the pH of the diluted solution?

**Answer**

Before we enter the parameters into the program, please recap: The dilution factor is

2000 mL / 40 mL = 50

Hence, the diluted solution contains exactly *(2500 mM / 50)* = *50 mM* H_{2}SO_{4}.

Switch to pure water (click on button *New*), then click on *Reac* and enter for “H2SO4” the value *50 mM*. Run the calculation with *Start*.

Instantly, the program will output the result (see right screenshot):

pH = 1.27

The diluted solution has pH 1.27.

**Note.** If you solve this task with a pocket calculator you would obtain *pH 1.3* as a raw approximation. The deviation from the correct result (pH 1.27) is caused by ionic activity corrections for non-ideal solutions.

*Problem 2: Mixing HCl with NaOH*

100 mL of 0.01 molar HCl is mixed with 50 mL of 0.05 molar NaCl. What is the pH of the final solution?

**Answer**

We begin with pure water (click on button *New*), then click on *Reac* and activate the checkbox *More Reactions*.

Before we enter the parameters, please recap: The final volume is 150 mL (= 100 mL + 50 mL) and, thus, the mixing solution contains:

( 100 mL / 150 mL ) × 10 mM = 6.67 mM HCl |

( 50 mL / 150 mL ) × 50 mM = 16.7 mM NaOH |

Run the calculation with *Start*. The result is shown on the right screenshot. So the answer is: The mixed solution has pH 11.94.

**Note.** If you solve this task with a pocket calculator you would obtain pH 12. The result calculated by *aqion*, however, is more accurate. The small deviation is caused by ionic activity corrections for non-ideal solutions.