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# ct whether an aqueous solution of each of the following will be acidic.docx

ct whether an aqueous solution of each of the following will be acidic, basic or neutral a. NH CH,CoO b. NH CN 11. A 0.52 5 g sample of a monoprotic acid is titrated with 0.125 M NaOH. The following titration curve is (14) obtained. a. The volume of base required to neutralize the acid b. Initial pH c. What is the molar mass of the acid? d. pH after 5.00 mL. of acid is added. e. pH at equivalence point f pH at one-half of the equivalence point. g. What is the pKa and K, of the acid? h. pH after adding 5.0 mL of base beyond the equivalence point. i. Calculate the initial pH from the K, value. j. Label the graph for buffer region. 14 12 10 8 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 mL. NaOH 0
Solution
Since ammonium acetate is a salt of weak acid and weak base, so it is often used to make buffer solution
for ammonium acetate (CH3COONH4) pKa=4.75 and pKb=4.75 too
NH4CN
This is a salt made up of salt of weak acid and weak base
neutral (NH4OH and HCN is weak acid and base
Thanks :)
.

ct whether an aqueous solution of each of the following will be acidic, basic or neutral a. NH CH,CoO b. NH CN 11. A 0.52 5 g sample of a monoprotic acid is titrated with 0.125 M NaOH. The following titration curve is (14) obtained. a. The volume of base required to neutralize the acid b. Initial pH c. What is the molar mass of the acid? d. pH after 5.00 mL. of acid is added. e. pH at equivalence point f pH at one-half of the equivalence point. g. What is the pKa and K, of the acid? h. pH after adding 5.0 mL of base beyond the equivalence point. i. Calculate the initial pH from the K, value. j. Label the graph for buffer region. 14 12 10 8 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 mL. NaOH 0
Solution
Since ammonium acetate is a salt of weak acid and weak base, so it is often used to make buffer solution
for ammonium acetate (CH3COONH4) pKa=4.75 and pKb=4.75 too
NH4CN
This is a salt made up of salt of weak acid and weak base
neutral (NH4OH and HCN is weak acid and base
Thanks :)
.

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