(a) What volume of 0.127 M KOH is needed toreach the equivalence point in the ti

Question

(a) What volume of 0.127 M KOH is needed toreach the equivalence point in the titration of 38.7 mL of0.0410 Mhypobromous acid? Write the name and thechemical formula for the salt formed during the titration.

b) What volume of0.330 M acetic acid is needed to reach theequivalence point in the titration of 31.2 mL of0.0480 MKOH? Write the name and the chemicalformula for the salt formed during the titration.
b) What volume of0.330 M acetic acid is needed to reach theequivalence point in the titration of 31.2 mL of0.0480 MKOH? Write the name and the chemicalformula for the salt formed during the titration.

Explanation / Answer

(a) Volume of KOH, V1 = ? Molarity of KOH M1 = 0.127 M . Volume of HBr , V2 = 38.7 mL Molarity, M2 = 0.041 M . M1 V1 = M2 V2 Volume of KOH, V1 = M2 V2 / M1                              = 0.041 M * 38.7 mL / 0.127 M                                = 12.49 mL . Name of the salt : Potassium bromide, KBr . (b)   Volume of acetic acid V1 = ?        M1 = 0.330 M Volume of KOH, V2 = 31.2 mL M2 = 0.048 M . V1 = M2 V2 / M1       = 0.048 M * 31.2 mL / 0.330 M     = 4.54 mL . Name of salt: Potassium acetate, CH3COOK

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