A student needed to standardize a solution of NaOH which was approximately 0.125 M. The student carefully prepared the titration setup, but after 25 mL of NaOH was added, what was the pH of solution if initially flask contains 25 mL of 0.250 M HCl solution?
http://www.jiskha.com/display.cgi?id=1287348651
To determine the pH of the solution after adding 25 mL of NaOH to 25 mL of 0.250 M HCl, we need to understand the reaction between NaOH and HCl and its impact on the pH.
NaOH (sodium hydroxide) is a strong base, while HCl (hydrochloric acid) is a strong acid. When sodium hydroxide reacts with hydrochloric acid, they undergo a neutralization reaction, resulting in the formation of water (H2O) and a salt (NaCl).
The balanced chemical equation for the reaction between NaOH and HCl is as follows:
NaOH + HCl → H2O + NaCl
In this reaction, NaOH and HCl react in a 1:1 stoichiometric ratio, meaning that for every mole of NaOH reacted, one mole of HCl is consumed.
Given that the initial volume of HCl is 25 mL and its concentration is 0.250 M, we can calculate the number of moles of HCl using the formula:
moles of HCl = concentration of HCl x volume of HCl
moles of HCl = 0.250 M x (25 mL / 1000) = 0.00625 moles
Since NaOH and HCl react in a 1:1 ratio, this means that 0.00625 moles of NaOH are required to neutralize the available HCl.
Now, let's consider the newly formed solution after adding 25 mL of NaOH. The total volume of the solution would be the sum of the initial volumes of NaOH and HCl, which is 50 mL.
To determine the new concentration of the solution, we can use the formula:
new concentration = moles / volume
new concentration of NaOH = 0.00625 moles / (50 mL / 1000) = 0.125 M
So, the concentration of NaOH in the solution is still approximately 0.125 M.
Now, let's talk about the pH of the solution. Since NaOH is a strong base, it dissociates completely in water, producing hydroxide ions (OH-) that increase the pH.
However, as we added an equal amount of HCl to the solution, which is a strong acid, it also dissociates completely, producing hydronium ions (H3O+) that decrease the pH.
But since NaOH and HCl react in a 1:1 stoichiometric ratio, the net effect on the pH will be determined by the difference between the pH of HCl and NaOH solutions.
In this case, the initial concentration of HCl is 0.250 M. Knowing that HCl is a strong acid, the pH of the HCl solution is equal to -log[HCl concentration].
pH of HCl solution = -log(0.250) = 0.602
For NaOH, we can assume that the initial pH is 14 - pH of HCl, since they react in equal stoichiometric quantities.
pH of NaOH solution = 14 - 0.602 = 13.398
So, after adding 25 mL of NaOH to 25 mL of 0.250 M HCl solution, the pH of the resulting solution would be approximately 13.398.