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 ad…
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Question “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 ad…”
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?
Answer
This problem was solved by stoichiometry or balance reaction. It is a concept that indicates the relationship between any chemical reaction’s reactant and product.
Reaction of acid-base neutralization
Acid-base neutralization is when an acid reacts with a base to create water or solvent.
Take, for example:
![Ca(OH), (aq) + 2HNO, (aq) → 2H20 (1) + Ca(NO2)2(aq)](https://drive.google.com/uc?id=1EwxHULlnHjVyOwk7bGPSC8yI1c9GXcUx&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
Molarity:
The molarity is the concentration of solute per liter in a solution. It is represented by
![](https://drive.google.com/uc?id=14jLBHRh9jlCB6aupyw1Krnzexi5NTfAB&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
. Molarity is the ratio of moles per liter solution. This formula explains it:
![moles of solutes
Molarity (M)=
liter of solution
mole
= mole LM](https://drive.google.com/uc?id=1ZdtNQF6ZNy7WhBDw3QIPDwNEKAfFUJIv&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
Titration of an acid and a base
Titration of an Acid with Base is a method to determine the amount of an acid or base by adding a calculated quantity.
Equivalent point:
An equivalent point is an acid-base titration point where acid and base are present in an equal amount.
Indicator:
An indicator of acid-base, or color change in indicator of acid-base titration, is a compound which changes color according
![](https://drive.google.com/uc?id=1nSIOvC1yScZb7-RH11CpNeIP-cd4Zvb2&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
changes to solution. The indicator of acid-base is a color that changes as the solution becomes more acidic or less acidic.
Endpoint:
The end point of acid-base titration refers to the point at which acidic solutions become basic or acidic solutions. At this point, the indicator’s color changes. An indicator is used in acid-base Titration to determine the end point.
The following expression is used to calculate the number moles:
![Moles of solutes = Molarity (M)x liter of solution or volume
Amount in g
Number of moles =
Molar mass](https://drive.google.com/uc?id=105iauosv0HmC-N6Fq4NF5Byq_0mWvgF6&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
[Part a]
This is the balance chemical reaction between strong acid and strong bas:
![HCl(aq) + NaOH(aq) + NaCl(aq) + H2O(1)
Reactants
Products](https://drive.google.com/uc?id=1j1KYRZOa2ZJZ0NBlEFA7jpr33RDIlIim&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
The following steps will allow you to calculate the moles of acid or base:
![HCI:
Moles of solutes = Molarity (M)x liter of solution or volume
1.0L
= 0.200M x100. ml x-
1000 ml
= 0.02 Moles HCI](https://drive.google.com/uc?id=1EKx1H-KGDun6pxaYeRMuzUi6zBTrp-j_&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
![NaOH:
Moles of solutes = Molarity (M)x liter of solution or volume
1.0 L
= 0.250M X 50. ml x
1000 ml
= 0.0125 Moles NaOH](https://drive.google.com/uc?id=1UDUcv7yrlLv-iIusVh32TCENM49VtQZX&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
The balance reaction of 1 mole
![](https://drive.google.com/uc?id=1_v3UIIU0WrHh4cL8iDKtSL9AfUHsMm7z&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
to 1 mole $9$ is called the “balance reaction”. Accordingly
![0.0125 Moles NaOH](https://drive.google.com/uc?id=1FcDC0-RVKAeX2EAlTvxpp7g0Imn-BTvJ&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
![](https://drive.google.com/uc?id=1QPArKI1HYEPL9GC_EJrx7XG2x--eKL0v&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
moles, which were calculated as follows:
![Total mole of HCI - used mole HCI = 0.02 Moles HCI-0.0125 Moles HCI
=0.0075 Moles HCI](https://drive.google.com/uc?id=1GeFhA7AyC1eDQNgIejU5_Y7dl8bGXGZr&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
The total volume of reaction is the volume of
![](https://drive.google.com/uc?id=1VIOpPvKzNw5Y6K63dSTbx_rJmLjS-W9v&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
+ the volume of
![NaOH](https://drive.google.com/uc?id=1iaZr39uPtGFalh1YMuJXCtK4Pq67a-mW&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
![100. ml+50. ml=150. ml](https://drive.google.com/uc?id=1426tLUsv6fu0KLIIO9ht4fVXeCmjE93_&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
Calculate the new molarity for
![](https://drive.google.com/uc?id=1VIOpPvKzNw5Y6K63dSTbx_rJmLjS-W9v&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
by following these steps:
![moles of solutes
Molarity (M)=
liter of solution
150 ml
0.0075 Moles HCI
-= 0.05 M HCI
. 1.0L
1000 ml](https://drive.google.com/uc?id=1BOAkWhj2_bsVoGWCYMWgaFjCRz_SyOyS&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
![](https://drive.google.com/uc?id=1_v3UIIU0WrHh4cL8iDKtSL9AfUHsMm7z&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
refers to a strong acid. Therefore, the molarity
![](https://drive.google.com/uc?id=1_v3UIIU0WrHh4cL8iDKtSL9AfUHsMm7z&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
equals the hydrogen ion concentration.
![](https://drive.google.com/uc?id=1pGhTVjHQ9zIKStRsXjx40qBJNnIKz_zX&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
determines the nature of the solution. It is:
![pH = -log,0 4,01
=-log,[0.05]
=-(-1.30)
=1.30](https://drive.google.com/uc?id=1arbkbcT_bp1_UHFf0nAbMCI_5EFHZMpJ&export=download/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-ad.png?x-oss-process=image/resize,w_560/format,webp)
[Part a]
Part of
Ans: Part A
This pH is the pH of the solution after adding 50.0ml base.
Conclusion
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