Question
A 20.0-mL sample of a 0.125 M diprotic acid (H2A) solution is titratedwith 0.1019 M KOH. The acid ionization constants for theacid are Ka1 = 5.2 * 10-5 and Ka2 = 3.4 * 10-10. At what addedvolume of base does each equivalence point occur?
Step 1
We can do this by multiplying the volume of the solution by the molarity of the acid. \[ moles_{H2A} = volume_{H2A} \times molarity_{H2A} \] \[ moles_{H2A} = 20.0 mL \times 0.125 M \] \[ moles_{H2A} = 0.0025 moles \] Show more…
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A 20.0-mL sample of a 0.125 M diprotic acid (H2A) solution is titrated with 0.1019 M KOH. The acid ionization constants for the acid are Ka1 = 5.2 * 10 - 5 and Ka2 = 3.4 * 10 - 10. At what added volume of base does each equivalence point occur?
A $20.0 \mathrm{~mL}$ sample of a $0.125 \mathrm{M}$ diprotic acid $\left(\mathrm{H}_{2} \mathrm{~A}\right)$ solution is titrated with $0.1019 \mathrm{M}$ KOH. The acid ionization constants for the acid are $K_{a_{1}}=5.2 \times 10^{-5}$ and $K_{\mathrm{a}_{2}}=$ $3.4 \times 10^{-10}$. At what added volume of base does each equivalence point occur?
21.0 mL of 0.125 M diprotic acid (H2A) was titrated with 0.1021 M KOH. The acid ionization constants for the acid are Ka1=5.2×10?5 and Ka2=3.4×10?10. At what added volume of base does the first equivalence point occur?
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