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Amanda [17]
3 years ago
9

For which type of titration will the ph be acidic at the equivalence point?

Chemistry
2 answers:
ANTONII [103]3 years ago
8 0
It is 
<span>b. strong acid vs. weak base.</span>
Ira Lisetskai [31]3 years ago
5 0

Equivalence point in an acid base titration is the point when there are equal moles of acid and base in the solution.

The pH at the equivalence point of a strong acid and a strong base will be 7 as the solution will be neutral. The conjugate acid and the conjugate base of a strong base and strong acid will be neutral.

HCl(aq) + NaOH(aq)---> NaCl(aq) + H_{2}O(l)

The pH at the equivalence point of a titration of weak acid and a strong base will be slightly basic as the conjugate base of the weak acid formed in the titration will be basic.

CH_{3}COOH (aq) + NaOH (aq)--> CH_{3}COONa(aq) + H_{2}O(l)

The pH at the equivalence point of a titration of strong acid and a weak base will be slightly acidic because the conjugate acid of the weak base is slightly acidic.

HCl(aq) + NH_{3}(aq) ---> NH_{4}^{+}(aq) + Cl^{-}(aq)

The pH at the equivalence point of the titration of a weak acid and a weak base is 7. But the pH changes are not sharp as both the acid and base are weak.

CH_{3}COOH (aq)+NH_{3}(aq) -->CH_{3}COONH_{4}(aq)

Therefore, the correct answer is B: Strong acid versus weak base

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What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?
Helen [10]
The balanced equation for the above reaction is as follows;
2HCl + K₂SO₃ ---> 2KCl + H₂O + SO₂
stoichiometry of HCl to SO₂ is 2:1
number of moles of HCl reacted - 15.0 g / 36.5 g/mol = 0.411 mol 
according to molar ratio 
number of SO₂ moles formed - 0.411 mol /2 = 0.206 mol
since we know the number of moles we can find volume using ideal gas law equation 
PV = nRT
where
 P - pressure - 1.35 atm x 101 325 Pa/atm = 136 789 Pa 
V - volume 
n - number of moles - 0.206 mol 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation 

136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K 
V = 4.07 L 
volume of SO₂ formed is 4.07 L


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<h3>What is supersaturation?</h3>

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Learn more about supersaturation

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