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aleksley [76]
3 years ago
7

Some students performed a titration between 20.0 mL of 0.5 M hydrochloric acid and 1.0 M potassium hydroxide solution. The stude

nts collected data and plotted the graph below. Which statement correctly explains the reaction at point D?
OPTION A) All hydroxide ions have reacted. There is no excess of hydroxide ions at this point.
OPTION B) The volume of base that has been added is equal to the volume of acid in the flask; this helps in balancing the ions present, making the pH of the solution neutral.
OPTION C) All hydrogen ions and all hydroxide ions have reacted to produce water, and so neither ion remains free in solution.
OPTION D) There are extra hydrogen ions in solution. As the base is added, the pH increases exponentially.

Chemistry
2 answers:
RideAnS [48]3 years ago
7 0

Answer:

OPTION C) All hydrogen ions and all hydroxide ions have reacted to produce water, and so neither ion remains free in solution.

Explanation:

The PH reading on the left axis of the graph shows a value of 7.

A value of 7 on the PH scale implies that equal number of hydrogen ions completely reacted with hydroxyl ions to produce water i.e there is no excess of hydrogen and hydroxyl ion remaining in solution.

For values greater than 7, a basic solution is formed and it signifies the presence of excess hydroxyl ion. If the value is less than 7, there is more hydrogen ions in the solution formed and it is said to be acidic.

At point 7, the hydrogen and hydroxyl ions are equal and completely neutralize out one another.

Note: The acid solution would require a base volume of 20mL to be completely neutralized according to the plot. If it is less, the PH shifts to the left and the solution becomes acidic. If it more, the solution becomes basic and the PH shifts rightwards.

kompoz [17]3 years ago
6 0

C on e 2020

extra words for 20 letters

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C₅H₁₀O₅

Explanation:

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(b) Mass of H

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2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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