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Calculate the ph of the resulting solution if 34.0 ml of 0.340 m hcl(aq is added to

Timothy Norman

in Chemistry

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Jessie Thompson on August 2, 2018

According to the reaction equation:HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (a) (Part 1):first, we need to get the moles of HCl = molarity * volume = 0.34 m * 0.034 = 0.01156 molthen, we need moles of NaOH = molarity * volume = 0.34 * 0.039 = 0.01326 molwhen NaOH in 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 = 0.0017 mol / 0.073 L = 0.0233 M∴ POH = -㏒[OH-] = -㏒0.0233 = 1.63 ∴ PH = 14 - 1.63 = 12.37 (b) part 2:as we have moles HCl = molarity * volume = 0.34 * 0.034 = 0.01156 mol so moles of NaOH = molarity * volume = 0.39 * 0.044 = 0.01716 molNaOH remaining = 0.01716 - 0.01156 = 0.0056 molwhen 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


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