Chemistry
TiffanyBarriault890
22

What volume of a 3.5 M HCL is required to completely neutralize 50.0 ml of a 2.0 M NaOH

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(1) Answers
KeriDrungo387

NaOH: V=50mL=0,05L Cm = 2M n = Cm*V = 0,05L * 2M = 0,1mol HCl         +    NaOH ⇒ NaCl + H₂O 1mol        :     1mol 0,1mol     :     0,1mol HCl: n = 0,1mol Cm = 3,5M         n             0,1mol V=---------- =   -----------        ≈ 0,029dm³ = 29mL of 3,5M HCl        Cm           3,5 mol/dm³

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