Pka=pH when[HA]=[A- ]. What is [A-]/[HA] WHEN PH=PKA+1 AND WHEN pH=pka-1
pH = pKa + log (A^-)/(HA)
When (A^-)/(HA) = 1, log 1 = 0 and pH = pKa.
When pH = pKa + 1, (A^-)/(HA) = 10 BECAUSE log 10 = 1.
I'll leave the pKa-1 for you to do.
When pH = pKa - 1, (A^-)/(HA) = 0.1 BECAUSE log 0.1 = -1.
To determine the ratio [A-]/[HA] when pH = pKa + 1 and pH = pKa - 1, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
1. When pH = pKa + 1:
Substitute the value of pH into the equation:
pKa + 1 = pKa + log([A-]/[HA])
Simplify the equation:
1 = log([A-]/[HA])
To remove the logarithm, we can rewrite the equation in exponential form:
10^1 = [A-]/[HA]
Therefore, [A-]/[HA] = 10.
2. When pH = pKa - 1:
Substitute the value of pH into the equation:
pKa - 1 = pKa + log([A-]/[HA])
Simplify the equation:
-1 = log([A-]/[HA])
Again, rewrite the equation in exponential form:
10^-1 = [A-]/[HA]
Therefore, [A-]/[HA] = 0.1.
In summary:
- When pH = pKa + 1, [A-]/[HA] = 10.
- When pH = pKa - 1, [A-]/[HA] = 0.1.