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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 molarity of KOH?​

Respuesta :

guevarajuan

Answer:

1.4 M

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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