A solution is prepared by mixing 0.01 mol each of H2CO3, NaHCO3, Na2CO3, and NaOH in 100 mL of water. pH of the resulting solution is _______.[Given: pKa1 and pKa2 of H2CO3 are 6.37 and 10.32, respectively; log 2 = 0.30]
Correct Answer :
` and ``. 2. Formatting rule: Separate paragraphs with `...` and use `` for line breaks within a paragraph. To prevent vertical cramping, place line containing `...` block in its own paragraph `...` or `` before and after. 3. MathML & Symbol rule: Output math using HTML / and MathML . No LaTeX ($..$). Do NOT use HTML hex or decimal entities for minus/arrows. Use raw unicode characters. Never use `display="block"` in `` tags, omit display attribute or set `display=""`. 4. State correct answer at beginning. Let's write out the MathML cleanly. Paragraph 1: The correct answer is 10.02. Paragraph 2: To find the pH of the solution, we first need to determine the chemical reactions taking place between the solute components: H2CO3 (a weak acid), NaHCO3 (supplying HCO3-), Na2CO3 (supplying CO32-), and NaOH (a strong base supplying OH-). Paragraph 3: Initial moles of each component present in 100 mL of solution: • Moles of H2CO3 = 0.01 mol • Moles of HCO3- = 0.01 mol • Moles of CO32- = 0.01 mol • Moles of OH- = 0.01 mol Paragraph 4: The strong base, OH-, will react completely with the strongest acid present, which is carbonic acid (H2CO3): H2CO3+OH-→HCO3-+H2O Wait, forbidden HTML entities rule: "VERY IMPORTANT: Do NOT use HTML hex or decimal entities (like − or −) for symbols like minus or arrows. Use standard raw unicode characters like -, +, =, ⇒ directly." So use `→` directly or `⇒` directly instead of `→`. Let's fix that: `→` Paragraph 5: Since 0.01 mol of OH- reacts completely with 0.01 mol of H2CO3: • Moles of H2CO3 remaining = 0.01 - 0.01 = 0 mol • Moles of OH- remaining = 0 mol • Moles of HCO3- produced = 0.01 mol Paragraph 6: Now, calculate the total moles of each species remaining in the solution: • Total moles of HCO3- = Initial + Produced = 0.01 mol + 0.01 mol = 0.02 mol • Total moles of CO32- = 0.01 mol Paragraph 7: The resulting mixture consists of a weak acid (HCO3-) and its conjugate base (CO32-), which forms an acid buffer system governed by the second dissociation constant (pKa2) of carbonic acid: HCO3-⇌H++CO32- Wait, raw unicode symbol for equilibrium: `⇌`. Paragraph 8: Using the Henderson-Hasselbalch equation for the buffer solution: pH=pKa2+log[CO32-][HCO3-] Paragraph 9: Substituting the values into the buffer formula: pH=10.32+log0.010.02 pH=10.32+log12 pH=10.32-log2 Paragraph 10: Given log 2 = 0.30: pH=10.32-0.30=10.02 Paragraph 11: Thus, the pH of the resulting solution is 10.02. Everything looks completely aligned with all guidelines! Let's output. 10.02
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The correct answer is 10.02.
To find the pH of the solution, we first need to determine the chemical reactions taking place between the solute components: H2CO3 (a weak acid), NaHCO3 (supplying HCO3-), Na2CO3 (supplying CO32-), and NaOH (a strong base supplying OH-).
Initial moles of each component present in 100 mL of solution: • Moles of H2CO3 = 0.01 mol • Moles of HCO3- = 0.01 mol • Moles of CO32- = 0.01 mol • Moles of OH- = 0.01 mol
The strong base, OH-, will react completely with the strongest acid present, which is carbonic acid (H2CO3):
H2CO3+OH-→HCO3-+H2O
Since 0.01 mol of OH- reacts completely with 0.01 mol of H2CO3: • Moles of H2CO3 remaining = 0.01 - 0.01 = 0 mol • Moles of OH- remaining = 0 mol • Moles of HCO3- produced = 0.01 mol
Now, calculate the total moles of each species remaining in the solution: • Total moles of HCO3- = Initial + Produced = 0.01 mol + 0.01 mol = 0.02 mol • Total moles of CO32- = 0.01 mol
The resulting mixture consists of a weak acid (HCO3-) and its conjugate base (CO32-), which forms an acid buffer system governed by the second dissociation constant (pKa2) of carbonic acid:
HCO3-⇌H++CO32-
Using the Henderson-Hasselbalch equation for the buffer solution:
pH=pKa2+log[CO32-][HCO3-]
Substituting the values into the buffer formula:
pH=10.32+log0.010.02
pH=10.32+log12
pH=10.32-log2
Given log 2 = 0.30:
pH=10.32-0.30=10.02
Thus, the pH of the resulting solution is 10.02.
Solution :
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To determine the pH of the resulting solution, we must analyze the neutralization reactions occurring when the given components are mixed together in water.
The initial amount of each component added to the solution is: • Moles of H2CO3 = 0.01 mol • Moles of NaHCO3 (supplying HCO3-) = 0.01 mol • Moles of Na2CO3 (supplying CO32-) = 0.01 mol • Moles of NaOH (supplying strong base OH-) = 0.01 mol
The strong base, OH-, reacts completely with the strongest acid present in the solution, which is carbonic acid (H2CO3):
Since exactly 0.01 mol of OH- is available, it completely neutralizes all 0.01 mol of H2CO3: • Moles of H2CO3 remaining = 0.01 - 0.01 = 0 mol • Moles of OH- remaining = 0 mol • Moles of HCO3- formed from reaction = 0.01 mol
After the reaction completes, the total amounts of chemical species present in the solution are: • Total moles of HCO3- = (Initial moles) + (Moles formed) = 0.01 mol + 0.01 mol = 0.02 mol • Total moles of CO32- = 0.01 mol
The solution now contains a mixture of the weak acid HCO3- and its conjugate base CO32-, forming an acidic buffer solution governed by the second ionization equilibrium of carbonic acid:
Using the Henderson-Hasselbalch equation for a buffer solution:
Substitute the given pKa2 = 10.32 and the molar amounts into the equation:
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