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14 October, 19:12

Calculate the ph of the resulting solution if 27.0 ml of 0.270 m hcl (aq is added to 37.0 ml of 0.270 m naoh (aq

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  1. 14 October, 19:16
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    The reaction between NaOH and HCl is as follows

    NaOH + HCl - - - > NaCl + H₂O

    for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.

    Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol

    number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol

    HCl reacts with NaOH in 1:1 molar ratio

    Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol

    total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L

    Since there's excess OH⁻ ions, we can calculate pOH value first

    pOH = - log [OH⁻]

    [OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L

    pOH = - log (0.042 M)

    pOH = 1.37

    by knowing pOH we can calculate pH using the following equation;

    pH + pOH = 14

    pH = 14 - 1.37

    pH = 12.63
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