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10 January, 11:38

The enthalpy change when a strong acid is neutralized by strong base is - 56.1 kj/mol. if 135 ml of 0.450 m hi at 23.15°c is mixed with 145 ml of 0.500 m naoh, also at 23.15°c, what is the maximum temperature reached by the resulting solution? (assume that there is no heat loss to the container, that the specific heat of the final solution is 4.18 j/g·°c, and that the density of the final solution is that of water.)

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  1. 10 January, 13:37
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    First, we have to get the moles of HI = molarity * volume L

    = 0.45 m * 0.135 L

    = 0.06075 moles

    then the moles of NaOH = molarity * volume L

    = 0.5 m * 0.145 L

    = 0.0725 moles

    when we have the reaction equation is:

    HI + NaOH → H2O + NaI

    and HI is the limiting reactant

    when Q = molar enthalpy change * moles HI

    = 56100 j / mol * 0.06075 mol

    = 3408 j

    when Q = M * Cp * ΔT

    M is the mass & Cp is the specific heat & ΔT is difference in temperature

    we need to get the mass (m), when the total volume = 135 ml + 145 ml

    =280 ml

    ∴ Mass (m) = volume * density

    = 280 * 1 = 280 g

    so by substitution:

    3408 = 280 g * 4.18 * ΔT

    ∴ΔT = 2.9 °C

    ∴ (Tf - Ti) = 2.9 °C

    when Ti = 23.15°C

    ∴Tf = 23.15 + 2.9

    = 26.05 °C
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