HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06 Solution HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06.
![HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added
10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] =
3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol)
n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 =
0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH
n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/
0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol)
n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] =
Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06
Solution
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added
10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] =
3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol)
n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 =
0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH
n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/
0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] =
7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol)
n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] =
Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06](https://image.slidesharecdn.com/hclo3naohnaclo3h2onhclo30-230409001440-b67dc782/85/HClO3-NaOH-NaClO3-H2O-n-HClO3-0-10-04-4-pdf-1-320.jpg)
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HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
- 1. HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06 Solution HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06