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7 March, 01:34

Assuming the final solution will be diluted to 1.00 l, how much more hcl should you add to achieve the desired ph?

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  1. 7 March, 02:37
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    H = 2.7, therefore amount of H + needed is 10^-2.7 M

    u haf 80 ml of hcl and 90 ml of naoh left, therefore 20 ml of hcl used, and 10ml of naoh used.

    mols of H + = 0.02 x 7 x 10^-2 = 1.4 x 10^-3

    mols of OH - = 0.01 x 5 x 10^-2 = 5 x 10^-4

    H + and OH - neutralise each other, so remaining mols of H + = 9 x 10^-4

    u need 10^-2.7 mols of H+, so 10^-2.7 - 9 x 10^-4 = 0.001095 mol

    vol of HCL needed = 0.001095 / 7 x 10^-2 = 0.0156 L = 15.6 mL
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