QuestionWhich reactions are neutralized by of and identify the best indicator for acetic acid?
Studdy Solution
STEP 1
Assumptions1. We are dealing with neutralization reactions, which are reactions between an acid and a base that produce a salt and water.
. The reactions are assumed to go to completion.
3. The number of moles of acid and base react according to their stoichiometry in the balanced chemical equation.
4. The pH indicators are assumed to change color at their typical pH ranges.
5. The pOH is the negative logarithm (base10) of the concentration of hydroxide ions.
6. An acidic solution has a higher concentration of hydronium ions than hydroxide ions.
STEP 2
First, let's solve problem7. We need to find which acid will be neutralized by4 moles of NaOH. The key here is to understand the stoichiometry of the reaction between the acid and the base. For a monoprotic acid (an acid that donates one proton per molecule), one mole of NaOH will neutralize one mole of the acid. For a diprotic acid (an acid that donates two protons per molecule), one mole of NaOH will neutralize half a mole of the acid.
STEP 3
HBr is a monoprotic acid, so2 moles of HBr will require2 moles of NaOH for neutralization.
STEP 4
H2Cr4 is a diprotic acid, so2 moles of H2Cr4 will require4 moles of NaOH for neutralization.
STEP 5
H3PO4 is a triprotic acid, so2 moles of H3PO4 will require moles of NaOH for neutralization.
STEP 6
HF is a monoprotic acid, so2 moles of HF will require2 moles of NaOH for neutralization.
STEP 7
Therefore,4 moles of NaOH will neutralize2 moles of H2Cr4. So, the answer to problem7 is (b).
STEP 8
For problem8, we need to choose the best pH indicator for the titration of acetic acid with NaOH. Acetic acid is a weak acid and NaOH is a strong base, so the equivalence point will be in the basic pH range.
STEP 9
Methyl orange changes color in the pH range3. to4.4, which is too low for our titration.
STEP 10
Methyl red changes color in the pH range4.4 to6.2, which is still too low for our titration.
STEP 11
Bromthymol blue changes color in the pH range6.0 to7.6, which is closer but still not ideal for our titration.
STEP 12
Phenolphthalein changes color in the pH range8.2 to10.0, which is ideal for our titration.
STEP 13
Therefore, the best choice of pH indicator for this titration is phenolphthalein. So, the answer to problem8 is (d).
STEP 14
For problem9, we need to find which solution has the lowest pOH. Since pOH is the negative logarithm of the concentration of hydroxide ions, the solution with the highest concentration of hydroxide ions will have the lowest pOH.
STEP 15
NaNO3 is a neutral salt, so its solutions do not affect the pH or pOH. Therefore, the pOH of0. M NaNO3 and M NaNO3 will be the same and equal to the pOH of pure water, which is7 at25 degrees Celsius.
STEP 16
NaNO2 is a basic salt, so its solutions will have a lower pOH than pure water. The pOH will be lower for a higher concentration of the salt.
STEP 17
Therefore, the solution with the lowest pOH is2 M NaNO2. So, the answer to problem9 is (c).
STEP 18
For problem10, we need to find which statement is true for an acidic solution at all temperatures.
STEP 19
In an acidic solution, the concentration of hydronium ions is greater than the concentration of hydroxide ions. So, statement (a) is true.
STEP 20
In an acidic solution, the concentration of hydronium ions is greater than x10^-7 M. So, statement (b) is true.
STEP 21
In an acidic solution, the concentration of hydroxide ions is less than1 x10^-7 M. So, statement (c) is true.
STEP 22
Therefore, all statements are true for an acidic solution at all temperatures. So, the answer to problem10 is (d).
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