Calculate the pH of a 25 mL sample of distilled water after the addition of 1 mL, 2 mL, 3 mL, 4 mL, and 5 mL of NaOH and HCL. (0.1 M HCl and NaOH)
In total, you need to show 10 calculations.
See your post below.
To calculate the pH of a solution, we need to know its concentration of hydrogen ions (H+). In the case of distilled water, it contains pure H2O molecules and a few auto-dissociated ions.
Since distilled water is neutral, the concentration of H+ ions is 1 x 10^-7 M (at 25°C). The pH of neutral water is calculated using the formula:
pH = -log[H+]
Now let's calculate the pH after the addition of NaOH and HCl solutions:
1. After adding 1 mL of 0.1 M NaOH:
The concentration of NaOH added is 0.1 M x (1 mL / 25 mL), which is 0.004 M.
NaOH is a strong base that dissociates completely, so we can assume that the resulting OH- concentration is 0.004 M.
The concentration of H+ ions in the water will decrease, as OH- ions react with them to form water:
New concentration of H+ ions = initial concentration - concentration of OH- ions
= 1 x 10^-7 M - 0.004 M
Calculate the new pH using the formula:
pH = -log[H+]
2. After adding 1 mL of 0.1 M HCl:
The concentration of HCl added is 0.1 M x (1 mL / 25 mL), which is 0.004 M.
HCl is a strong acid that dissociates completely, so we can assume that the resulting H+ concentration is 0.004 M.
The concentration of H+ ions in the water will increase due to the addition of HCl:
New concentration of H+ ions = initial concentration + concentration of HCl
= 1 x 10^-7 M + 0.004 M
Calculate the new pH using the formula:
pH = -log[H+]
3. Repeat the same process for the other volumes of NaOH and HCl added (2 mL, 3 mL, 4 mL, and 5 mL), using their respective concentrations.
By following this procedure, you can calculate the pH after the addition of each NaOH and HCl solution, totaling 10 calculations.