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SVEN [57.7K]
2 years ago
11

an acid based titration was performed. it took 27.45 mL of the base, KOH, to titrate 3.115 g HBr, the acid. what was the molarit

y of KOH?​
Chemistry
1 answer:
sveticcg [70]2 years ago
7 0

Answer:

<u>1.4 M</u>

Explanation:

n(HBr)=3.115/81

so, 3.115/81=0.0385mol

according to the reaction, n(HBr)=n(KOH)=0.038 mol

C(KOH)=n/V=0.0385/0.02745

0.0385/0.02745 =1.4 M

in short the answer is 1.4 M (molarity)

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Answer:

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case e: 50.0 mL of 1.00 M NaOH

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case f: 200 mL of 1.00 M NaOH

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N_1V_1 hence it is after the eqivalence point

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