Two weak acids, a and b, have pka values of 4 and 6, respectively. which statement is true? a) acid a dissociates to a greater extent in water than acid b. b) for solutions of equal concentration, acid b will have a lower ph. c) b is the conjugate base of a. d) acid a is more likely to be a polyprotic acid than acid b. e) the equivalence point of acid a is higher than that of acid b

A) Acid A dissociates to a greater extent in water than acid B

Explanation:

A) Acid A dissociates to a greater extent in water than acid B

We are given the pKa for both acids, and we know

pKa = - log Ka

Taking antilog to both sides of the equation we can solve for Ka

⇒ -pKa = log Ka

-antilog pKa = Ka

10 ^-pka = Ka

So ka for acid A = 10⁻⁴

and

ka for acid B = 10⁻⁶

True the equilibrium constant for acid A is greater, so it dissociates more.

B) For solutions of equal concentration, acid B will have a lower pH.

We know the stronger acid is A, and it dissociates more. Since pH is the negative log of H₃O⁺ concentration, it follows that at equal concentrations the acid A will have at equilibrium a greater [H₃O⁺] and hence a lower pH

C) B is the conjugate base of A

False:

If B were the conjugate base of A, its  Kb would have been given by:

Ka x Kb = Kw

Kb = 10⁻¹⁴/10⁻⁶ = 10⁻⁸ for the conjugate base of acid B

Kb = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ for the conjugate base of acid A

which are not equal.

D) Acid A is more likely to be a polyprotic acid than acid B.

False

Just having the pkas for both acids one cannot know if any of the acids is polyprotic. We will need the formula for the acids.

E) The equivalence point of acid A is higher than that of acid B

False

The equivalence point depends on the the concentration of the acids  and their volumes.

The equivalence point is reached in the titration when the number of equivalents of base equals the number of equivalents of acid:

# equivalents acid = # equivalents of base  @ end point

and

# equivalents acid = Molarity of acid x Volume of acid

answer: mercury

explanation:

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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