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12 May, 16:18

Given the NaOH density = 1.0698g mL^-1, How many mL of. 71M HCl would be required to neutralize 11.33g of 2.03 M NaOH?

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  1. 12 May, 18:35
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    The volume in ml of 0.71 M HCl that would be required to neutralize 11.33 g of 2.03 M NaOH is 30.3 ml

    calculation

    Step 1 : write equation for reaction

    NaOH + HCl →NaCl + H₂O

    Step 2:find the volume of NaOH

    volume=mass / density

    = 11.33 g / 1.0698 g/ml = 10.59 ml

    Step 3: find the moles of NaOH

    moles = molarity x volume in liters

    molarity = 2.03 M=2.03 mol/l

    volume in liters = 10.59/1000 = 0.0106 L

    moles = 2.03 M x 0.0106 L = 0.0215 moles

    step 4: use the mole ratio to determine the moles of HCl

    from equation in step 1, NaOH:HCl is 1:1 therefore the moles of HCl is also 0.0215 moles

    Step 5: find volume of HCl

    volume = moles / molarity

    molarity = 0.71 M = 0.71 mol/l

    =0.0215 moles / 0.71 mol/l=0.0303 L

    into ml = 0.0303 x 1000=30.3 ml
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