These are two questions and two answers.
Answer:
Explanation:
Question 1: What is the pH of the solution after 50.0ml of base has been added?
1) Titration: is the use of a neutralization reaction between an acid and a base to determine the concentraion of one of the solutions.
2) Neutralization reaction:
The neutralization of a strong acid and a strong base is represented by:
3) HCl (strong acid) + NaOH (strong base)
4) Number of moles of each reactant:
a) 100.0 ml of 0.200 M HCl
b) 50.0 ml of 0.250 M NaOH
5) Limiting and excess reactants:
Since, the stoichiometry is 1 : 1, 0.0125 mol of NaOH will react with 0.0125 mol of HCl. This means that NaOH is the limiting reactant, and there will be an excess of 0.0200 mol - 0.0125 mol = 0.0075 mol of HCl.
6) pH
[H⁺] = 0.0075 mol (the same concentration of HCl left as excess because each mole of HCl ionizes into one mole of H⁺ and one mol of Cl⁻)
Answer: the pH of the solution after 50.0 ml of base has been added is 2.12.
Question 2.What is the pH of the solution at the equivalence point?
This is a theortical question.
The equivalence point of the titration of a strong acid with a strong base is the point where the number of moles of H⁺ from the acid equals the number of moles of OH⁻ base.
Hence, at the equivalence point, all the H⁺ and OH⁻ will be converted into H₂O and so the pH is the same pH of pure water.
Such pH is known as neutral pH and is equal to 7 (at 25°C)
So, the answer is that the pH of a the solution at the equivalence point is 7.
In this exercise we will have to analyze the ph of the given solutions:
1) [tex]pH: 13.0[/tex]
2) solution is neutral.
In this exercise, you must identify which components correspond to the concentrate and the volume:
[tex]100 ml of 0.200M HCl\\100 ml = V\\0.200M= C\\and\\50 ml of 0.250 M NaOH\\50 ml = V\\0.250M= C[/tex]
After that, using the formula below, we will start the calculations:
[tex]N= C*V\\N_{HCl} = (0.2M)(0.1)= 0.02 mol \\N_{NaOH}= (0.250M)(0.05)= 0.0125 mol NaOH[/tex]
So when writing the equation:
[tex]HCl+NaOH\rightarrow NaCl+H2O\\[/tex]
rewriting the NaOH concentration in terms of moles:
[tex]C=\frac{N}{V} \\C= \frac{7.5X1^{-3} }{0.15} = 0.05 M[/tex]
So when writing the equation:
[tex]HCl+H2O \rightarrow H3O^{+} + Cl^{-}[/tex]
So calculating your ph we will have:
[tex]pH: -log{H3O}\\= - log(0.05)\\=13.0\\[/tex]
2) So calculating the equivalent point will be:
[tex]N_{HCl}= N_{NaOH}[/tex]
[tex]PH= 7 \\[/tex]
It can be identified that the solution is neutral.
Learn more: brainly.com/question/491373
