In each of these strong-acid - strong-base titrations, determinethe volume of ti
ID: 681841 • Letter: I
Question
In each of these strong-acid - strong-base titrations, determinethe volume of titrant that would effect a neutralization. (a) 0.016 L of0.72 M HCl titrated with0.48 M NaOH1 L
(b) 49.3 mL of 0.75 M NaOH titrated with 0.68 M HCl
2 mL
(c) 21.3 mL of 0.97 M H2SO4 titratedwith 0.68 M KOH
3 mL
(d) 0.063 L of 0.27 M NaOH titrated with 0.22 M H2SO4
4 L (a) 0.016 L of0.72 M HCl titrated with0.48 M NaOH
1 L
(b) 49.3 mL of 0.75 M NaOH titrated with 0.68 M HCl
2 mL
(c) 21.3 mL of 0.97 M H2SO4 titratedwith 0.68 M KOH
3 mL
(d) 0.063 L of 0.27 M NaOH titrated with 0.22 M H2SO4
4 L
Explanation / Answer
Balanced chemical equation: HCl + NaOH ------------> NaCl +H2O a ) Number of moles of HCl = 0.016 L *0.72 M = 0.01152 moles Number of moles of NaOH = 0.48 M * x But at at neutralisation point both are equal, 0.01152 moles = 0.48 x M x = 0.01152 moles / 0.48 M = 0.024 L b) Number of moles of HCl = 0.68 M * x Number of moles of NaOH = 0.75 M * 0.0493L = 0.036975 moles But at at neutralisation point both are equal, 0.036975 moles = 0.68 x M x = 0.036975 moles / 0.68 M = 0.0543 L = 54.3 mL c) H2SO4 +2 KOH------------> K2SO4 + 2H2O H2SO4 KOH M1 = 0.97M M2 = 0.68 M V1 = 0.0213L V2 = ? n1 =2 n2 = 1 M1V1 / n1 = M2V2 / n2 V2 = M1V1 * n2/ n1M2 =0.97 M * 0.0213 L * 1 / 2 * 0.68 M = 0.0151 L d) H2SO4 +2 NaOH ------------> Na2SO4 +2 H2O H2SO4 NaOH M1 = 0.22M M2 = 0.27 M V1 = x V2 = 0.063 L n1 =2 n2 = 1 M1V1 / n1 = M2V2 / n2 V1 = M2V2 * n1/ n2M1 = 0.27 M * 0.063 L* 2 / 1 * 0.22 M = 0.154 L b) Number of moles of HCl = 0.68 M * x Number of moles of NaOH = 0.75 M * 0.0493L = 0.036975 moles But at at neutralisation point both are equal, 0.036975 moles = 0.68 x M x = 0.036975 moles / 0.68 M = 0.0543 L = 54.3 mL c) H2SO4 +2 KOH------------> K2SO4 + 2H2O H2SO4 KOH M1 = 0.97M M2 = 0.68 M V1 = 0.0213L V2 = ? n1 =2 n2 = 1 M1V1 / n1 = M2V2 / n2 V2 = M1V1 * n2/ n1M2 =0.97 M * 0.0213 L * 1 / 2 * 0.68 M = 0.0151 L d) H2SO4 +2 NaOH ------------> Na2SO4 +2 H2O H2SO4 NaOH M1 = 0.22M M2 = 0.27 M V1 = x V2 = 0.063 L n1 =2 n2 = 1 M1V1 / n1 = M2V2 / n2 V1 = M2V2 * n1/ n2M1 = 0.27 M * 0.063 L* 2 / 1 * 0.22 M = 0.154 L H2SO4 NaOH M1 = 0.22M M2 = 0.27 M V1 = x V2 = 0.063 L n1 =2 n2 = 1 M1V1 / n1 = M2V2 / n2 V1 = M2V2 * n1/ n2M1 = 0.27 M * 0.063 L* 2 / 1 * 0.22 M = 0.154 LRelated Questions
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