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spayn [35]
3 years ago
12

Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,

consider how the situation affects the calculation of molarity of NaOH and moles NaOH, which is calculated from the mass of KHP. For each answer, you must supply a clear explanation for your answer. Often times, students will restate the facts rather than explaining why they chose the answer they chose.
a. You add the weighed KHP to a flask containing a 60 mL of water rather than 50 mL of water. Explain. Your answer must contain a well thought out and clearly written explanation.
b. The buret is still wet with water on the inside when you add your NaOH solution Explain.
c. The KHP is wet when you weigh it. Explain.
d. You titrate past the equivalence point by 0.50 mL.
Chemistry
1 answer:
vodomira [7]3 years ago
7 0

Answer:

Explanation:

The result will be affected.

The mass of KHP  weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

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<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}

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\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}

Based on no. of iron reacted,  

\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)

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Explanation:

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=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺

Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸

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