Chemistry, 14.10.2019 00:30 angelisevega

100. ml of 0.200m hcl is titrated with 0.250m naoh.
1. what is the ph of the solution after 50.0ml of base has been added?
2.what is the ph of the solution at the equivalence point?

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Answer from: twistedhyperboles

These are two questions and two answers.

Question 1: the pH of the solution after 50.0 ml of base has been added is 2.12.Question 2: the pH os the solution at the equivalence point is 7.

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:

Acid + Base → Salt + Water

3) HCl (strong acid) + NaOH (strong base)

HCl + NaOH → NaCl + H₂O

4) Number of moles of each reactant:

a) 100.0 ml of 0.200 M HCl

M = n / V (in liter) ⇒ n = M × V = 0.100 liter × 0.200 M = 0.0200 mol

b) 50.0 ml of 0.250 M NaOH

M = n / V (in liter) ⇒ n = M × V = 0.0500 liter × 0.250 M = 0.0125 mol

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

pH = - log [H⁺]

[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⁻)

pH = - log (0.0075) = 2.12

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.

Answer from: dmccray3357

In this exercise we will have to analyze the ph of the given solutions:

1) pH: 13.0

2) solution is neutral.

In this exercise, you must identify which components correspond to the concentrate and the volume:

$100 ml of 0.200M HCl\\100 ml = V\\0.200M= C\\and\\50 ml of 0.250 M NaOH\\50 ml = V\\0.250M= C$

After that, using the formula below, we will start the calculations:

$N= C*V\\N_{HCl} = (0.2M)(0.1)= 0.02 mol \\N_{NaOH}= (0.250M)(0.05)= 0.0125 mol NaOH$

So when writing the equation:

$HCl+NaOH\rightarrow NaCl+H2O\\$

rewriting the NaOH concentration in terms of moles:

$C=\frac{N}{V} \\C= \frac{7.5X1^{-3} }{0.15} = 0.05 M$

So when writing the equation:

$HCl+H2O \rightarrow H3O^{+} + Cl^{-}$

So calculating your ph we will have:

$pH: -log{H3O}\\= - log(0.05)\\=13.0\\$

2) So calculating the equivalent point will be:

$N_{HCl}= N_{NaOH}$

$PH= 7 \\$

It can be identified that the solution is neutral.

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100. ml of 0.200m hcl is titrated with 0.250m naoh.
1. what is the ph of the solution after 50...

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