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11 June, 02:04

In a titration, 25.9 mL of 3.4 x10^-3 M Ba (OH) 2 neutralized 16.6 mL of HCL solution. What is the molarity of the HCL solution?

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  1. 11 June, 05:14
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    First, We have to write the equation for neutralization:

    Ba (OH) 2 + 2HCl → BaCl2 + 2H2O

    so, from the equation of neutralization, we can get the ratio between Ba (OH) 2 and HCl. Ba (OH) 2 : HCl = 1:2

    - We have to get the no. of moles of Ba (OH) 2 to do the neutralization as we have 25.9ml of 3.4 x 10^-3 M Ba (OH) 2.

    So no. of moles of Ba (OH) 2 = (25.9ml/1000) * 3.4x10^-3 = 8.8 x 10^-5 mol

    and when Ba (OH) 2 : HCl = 1: 2

    So the no. of moles of HCl = 2 * (8.8x10^-5) = 1.76 x 10^-4 mol

    So when we have 1.76X10^-4 Mol in 16.6 ml (and we need to get it per liter)

    ∴ the molarity = no. of moles / mass weight

    = (1.76 x 10^-4 / 16.6ml) * (1000ml/L) = 0.0106 M Hcl
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