Achieve W2 #10
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Achieve W2 #10
I was wondering how to approach this problem because I wasn't quite sure how to do it. Any help is appreciated! Thank you!
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Re: Achieve W2 #10
For Weak Base, When the pH of the solution is equal to the pKa of the conjugate acid (BH+), there are equal amounts of the weak base (B) and the conjugate acid (BH+) in solution.
At pH values below the pKa, the charged, protonated species of BH+ will become more predominant in solution.
At pH values above the pKa, the neutral, deprotonated species of B will become more predominant in solution.
At pH values below the pKa, the charged, protonated species of BH+ will become more predominant in solution.
At pH values above the pKa, the neutral, deprotonated species of B will become more predominant in solution.
Re: Achieve W2 #10
For weak acid:
HA(aq)+H2O(l)↽−−⇀A−(aq)+H3O+(aq)
pH equal pKa & pH less than pKa, it would be neutral. Because HA will become more predominant in solution.
pH greater than pKa, it would be charged. Because A- will become more predominant in solution.
For weak base:
B(aq)+H2O(l)↽−−⇀BH+(aq)+OH−(aq)
pH equal pKa & pH greater than pKa, it would be neutral. Because B will become more predominant in solution.
pH less than pKa, it would be charged. Because BH+ will become more predominant in solution.
HA(aq)+H2O(l)↽−−⇀A−(aq)+H3O+(aq)
pH equal pKa & pH less than pKa, it would be neutral. Because HA will become more predominant in solution.
pH greater than pKa, it would be charged. Because A- will become more predominant in solution.
For weak base:
B(aq)+H2O(l)↽−−⇀BH+(aq)+OH−(aq)
pH equal pKa & pH greater than pKa, it would be neutral. Because B will become more predominant in solution.
pH less than pKa, it would be charged. Because BH+ will become more predominant in solution.
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