In an acid-base titration, 10.0 mL (VB) of NH3 (Ka(NH4+) = 5.8×10-10) 0.18 M (CB) are titrated at the equivalence point (EP) with 10.0 ml (VA) of HCl 0.18 M (CA). Calculate (to two decimal places) the pH of the solution at the equivalence point.
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At the equivalence point, the amount of acid and base are equal. This means that all the NH3 has reacted with HCl to form NH4+. Show more…
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Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. $100.0 \mathrm{mL}$ of $0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)$ titrated with $0.10 \mathrm{M} \mathrm{NaOH}$ b. $100.0 \mathrm{mL}$ of $0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)$ titrated with $0.20 \mathrm{M} \mathrm{HNO}_{3}$ c. $100.0 \mathrm{mL}$ of $0.50 M \mathrm{HCl}$ titrated with $0.25 \mathrm{M} \mathrm{NaOH}$
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