Question

Part C
A volume of 40.0 mL of a 0.280 M HNO3 solution is titrated with 0.820 M KOH. Calculate the volume of KOH required to reach the equivalence point.
Express your answer to three significant figures, and include the appropriate units.

How many moles of Ba(OH)2 are present in 235 mL of 0.800 M Ba(OH)2?
Express your answer with the appropriate units.

Answer 1

Answer: - A The Balanced equation is HNO_3 + KOH rightarrow H_2 O + KNO_3 The equivalence point is the point at which equal moles of HNO_3 and KOH have been brought together. Since HNO_3 and KOH react in a one - to - one ratio then the moles of HNO_3 will equals the moles of KOH at the equivalence point. Here, we used the formula M_1 V_1 = M_2 V_2 (1) M_1 V_1 stands for HNO_3 where M_1 = 0.280 M V_1 = 40.0 ml and M_2 V_2 stands for KOH where, M_2 = 0.8202 M V_2 = x ml Put these values in the eqn. (1) we have, 0.280 M times 40.0 ml = 0.820 M times x ml = 0.280 times 40.0 = 0.820 times x x = 0.280 times 40.0/0.820 x 13.658 ml. Thus, 13.658 ml 0.820 M KOH is required to reach the equivalence point.\r\n < null >