2.A 15.0 ml sample of nitric acid solution istitration with a 0.960M barium hydr

Question

2.A 15.0 ml sample of nitric acid solution istitration with a 0.960M barium hydroxide until the end point. Based on the titration data, the concentration of nitric acid isdetermined to be 3.18M. Calculate the volume in millimetersof barium hydroxide used in this titration. Answer in unitsof mL

3.A 10.0 ml sample of 1.40M phosphoric acid isdiluted with 70.0 ml of deionized water and then completelyneutralized with 45.20ml of sodium hydroxide. What is theconcentration of the sodium hydroxide solution used in thisneutralization reaction? Answer in units of M.

Explanation / Answer

3Ca( OH ) 2 + 2H 3 PO 4 ------> Ca 3 ( PO 4 ) 2 + 6H2 O At nuetralization , NV = N ' V ' Given M = 0.85 M Normality,N =  Molarity * basicity                = 0.85 *3                = 2.49 N           V = 1L So, M ' V ' = No.of moles of Ca ( OH )2                    = n So, n = 2.49 * 1 =2.49 moles So, weight Ca ( OH ) 2 = n * Molecular weight of Ca( OH ) 2                                    = 2.49 * 74                                    = 184.26 g (2) . Given N = Molarity * acidity              = 0.96 * 2              = 1.92 N          N '=  Molarity * acidity                =3.18 * 1               = 3.18 N          V ' = 15mL          V = ? By the condition of neutrilization  N V = N ' V'                                                       V = 24.84 mL (3) . Given M = 1.4 M           V = 10mL           V ' =70 mL + 10 mL = 80 mL So, M ' = M V / V '            = 0.175 M Now the resulating 0.175 molar phosphoric acid is nuetralizdwith 45.2 mL of NaOH By the condition of neutralization , N" V " = N ' V'    here V " = 45.2 mL                  N ' = Molarity * basicity = 0.175 * 3 = 0.525N                  V ' = 10 + 70 = 80 mL                                                         N" = 0.9292 N Molarity = Normality / acidity = 0.9292 / 1 = 0.9292M Given N = Molarity * acidity              = 0.96 * 2              = 1.92 N          N '=  Molarity * acidity                =3.18 * 1               = 3.18 N          V ' = 15mL          V = ? By the condition of neutrilization  N V = N ' V'                                                       V = 24.84 mL (3) . Given M = 1.4 M           V = 10mL           V ' =70 mL + 10 mL = 80 mL So, M ' = M V / V '            = 0.175 M Now the resulating 0.175 molar phosphoric acid is nuetralizdwith 45.2 mL of NaOH By the condition of neutralization , N" V " = N ' V'    here V " = 45.2 mL                  N ' = Molarity * basicity = 0.175 * 3 = 0.525N                  V ' = 10 + 70 = 80 mL                                                         N" = 0.9292 N Molarity = Normality / acidity = 0.9292 / 1 = 0.9292M

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