As the number of effective collisions between reacting particles increases, the rate if a chemical reaction:
1 answer:
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According to the reaction equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
a) Part 1):
first, we need to get moles HCl = molarity * volume
= 0.34m * 0.034
= 0.01156 mol
then, we need moles of NaOH = molarity * volume
= 0.34 * 0.039
= 0.01326 mol
when NaOH excess:
∴ NaOH remaining = 0.01326 - 0.01156
= 0.0017 mol
when the total volume = 0.039 + 0.034
= 0.073 L
∴[OH] = moles / total volume
= 0.0017mol / 0.073 L
= 0.0233 M
∴ POH = -㏒[OH-]
= -㏒0.0233
= 1.63
∴ PH = 14- 1.63
= 12.37
b) part 2:
as we got moles HCl = molarity * volume
= 0.34 * 0.034
= 0.01156 mol
then moles NaOH = molarity * volume
= 0.39 * 0.044
= 0.01716 mol
NaOH remaining = 0.01716 - 0.01156
= 0.0056 mol
when the total volume = 0.034 + 0.044
= 0.078 L
∴[OH-] = 0.0056 / 0.078
= 0.0718 M
∴ POH = -㏒[OH-]
= -㏒0.0718
= 1.14
∴PH = 14 - POH
= 14 -POH
= 14 - 1.14
= 12.86
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
a) Part 1):
first, we need to get moles HCl = molarity * volume
= 0.34m * 0.034
= 0.01156 mol
then, we need moles of NaOH = molarity * volume
= 0.34 * 0.039
= 0.01326 mol
when NaOH excess:
∴ NaOH remaining = 0.01326 - 0.01156
= 0.0017 mol
when the total volume = 0.039 + 0.034
= 0.073 L
∴[OH] = moles / total volume
= 0.0017mol / 0.073 L
= 0.0233 M
∴ POH = -㏒[OH-]
= -㏒0.0233
= 1.63
∴ PH = 14- 1.63
= 12.37
b) part 2:
as we got moles HCl = molarity * volume
= 0.34 * 0.034
= 0.01156 mol
then moles NaOH = molarity * volume
= 0.39 * 0.044
= 0.01716 mol
NaOH remaining = 0.01716 - 0.01156
= 0.0056 mol
when the total volume = 0.034 + 0.044
= 0.078 L
∴[OH-] = 0.0056 / 0.078
= 0.0718 M
∴ POH = -㏒[OH-]
= -㏒0.0718
= 1.14
∴PH = 14 - POH
= 14 -POH
= 14 - 1.14
= 12.86