Question

In: Chemistry

1. If a small amount of strong base were added to a buffer made of a...

1. If a small amount of strong base were added to a buffer made of a weak acid HA and the lithium salt of its conjugate base, LiA, the pH does not change very much because....

a. No reaction occurs

b. The strong base reacts with A^- to give HA

c. Adding a strong base changes the pKa of the base

d. The strong base reacts with HA to give A^-

e. The strong base reacts with the Li⁺ to give LiOH

2) Which one of the following 0.10 M aqueous solutions has the highest pH?

a. Sodium perchlorate

b. Magnesium dichromate

c. Barium nitrate

d. Sodium sulfate

e. Iron (III) chloride

3) What is the equilibrium equation for the reaction 1.0 Molar solution of lithium dichromate with water?

a. Li₂Cr₂O₇(aq) + H2O(l) <-- --> 2 Li(aq)+ + C₂O₇^2- (aq)

b. Li(aq)⁺ + H₂O(aq) <-- --> LiH(aq) ⁺ + ^-OH(aq)

c. H₃O⁺ + ^-OH <-- --> 2 H₂O

d. Cr₂O₇^2-(aq)+ 2 H₂O(l) <-- --> H₂Cr₂O₇(aq) + 2 ^-OH(aq)

e. Cr₂O₇^2- (aq) + H₂O(l) <-- -->HCr₂O₇^- (aq) + ^-OH(aq)

Solutions

Expert Solution

1. If a small amount of strong base were added to a buffer made of a weak acid HA and the lithium salt of its conjugate base, LiA, the pH does not change very much because....

a. No reaction occurs

b. The strong base reacts with A- to give HA

c. Adding a strong base changes the pKa of the base

d. The strong base reacts with HA to give A-; since adding a base will increase conjguate base, and decrease weak acid

e. The strong base reacts with the Li+ to give LiOH

2) Which one of the following 0.10 M aqueous solutions has the highest pH? we need a basic soltion

a. Sodium perchlorate --> neutral

b. Magnesium dichromate MgCrO4 --> acidic CrO4 will take H+ ions, therefore this is the highest pH

c. Barium nitrate ---> acidic

d. Sodium sulfate --> neutral

e. Iron (III) chloride --> acidic, since Fe(OH)3 forms and H+ are free

3) What is the equilibrium equation for the reaction 1.0 Molar solution of lithium dichromate with water?

a. Li2Cr2O7(aq) + H2O(l) <-- --> 2 Li(aq)+ + C2O72- (aq)

b. Li(aq)+ + H2O(aq) <-- --> LiH(aq) + + -OH(aq)

c. H3O+ + -OH <-- --> 2 H2O

d. Cr2O72-(aq)+ 2 H2O(l) <-- --> H2Cr2O7(aq) + 2 -OH(aq)

e. Cr2O72- (aq) + H2O(l) <-- -->HCr2O7- (aq) + -OH(aq)

queen_honey_blossom answered 1 year ago

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