If someone could please help me with this i'd greatly appreciate it! I can't figure this one out and I've tried repeatedly.
Calculate the pH when 85.0 mL of 0.250 M HCl is mixed with 40.0 mL of 0.150 M Ca(OH)₂.
Chemistry Help Needed!!?
2020-04-03 12:48 pm
回答 (2)
2020-04-03 1:06 pm
HCl is a strong acid. Each mole of HCl dissociates to give 1 mole of H⁺ ions.
Moles of H⁺ ions added = (0.250 mol/L) × (85.0/1000 L) = 0.02125 mol
Ca(OH)₂ is a strong base. Each mole of Ca(OH)₂ dissociates to give 1 mole of OH⁻ ions.
Moles of OH⁻ ions added = (0.150 mol/L) × (40.0/1000 mL) × 2 = 0.012 mol
H⁺ + OH⁻ → H₂O
H⁺ is in excess.
Moles of H⁺ left unreacted = (0.02125 - 0.012) mol = 0.00925 mol
Volume of the final solution = (85.0 + 40.0) mL = 125.0 mL = 0.125 L
[H⁺] in the final solution = (0.00925 mol) / (0.125 L) = 0.074 M
pH = -log[H⁺] = -log(0.074) = 1.13
Moles of H⁺ ions added = (0.250 mol/L) × (85.0/1000 L) = 0.02125 mol
Ca(OH)₂ is a strong base. Each mole of Ca(OH)₂ dissociates to give 1 mole of OH⁻ ions.
Moles of OH⁻ ions added = (0.150 mol/L) × (40.0/1000 mL) × 2 = 0.012 mol
H⁺ + OH⁻ → H₂O
H⁺ is in excess.
Moles of H⁺ left unreacted = (0.02125 - 0.012) mol = 0.00925 mol
Volume of the final solution = (85.0 + 40.0) mL = 125.0 mL = 0.125 L
[H⁺] in the final solution = (0.00925 mol) / (0.125 L) = 0.074 M
pH = -log[H⁺] = -log(0.074) = 1.13
2020-04-03 12:51 pm
(85 mL * 0.25 M - 2 * 40mL * 0.15 M) / (85 mL + 40 mL) = 0.074 M HCl remaining
pH = 1.14 (decimal place wrong)
pH = 1.14 (decimal place wrong)
收錄日期: 2021-04-18 18:26:31
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