A student makes up 2 l of a 3. 0 m aqueous solution of HBr. to this solution she adds 1 l of a 4 m NaOH solution, then the pH of the resulting solution is 13.83.
Normality is defined as the ratio of equivalent mass of compound to the volume of solution in litre.
N1 is the normality of HBR = 3m
N2 is the normality of NaOH = 4m
V1 is the volume of the HBr = 2l
V2 is the volume of the NaOH = 1l
V is the total volume of solution = V1 + V2 = 2l +1l
= 3l
Normality of the solution = N
As we know that,
N1V1 - N2V2 = NV
3 × 2 – 4 × 1 = N × 3
6 – 4 = N – 3
N = 2/3 = 0.67 m
(OH-) = N = 0.67 m
Since normality of NaOH is greater than BR . Hence the resulting solution is alkaline.
pOH = -log[OH-]
POH = - log (0.67) = 0.17
As we know that,
PH + pOH = 14
pH = 14 – 0.17
= 13.83
Thus, we calculate that the pH of the resulting solution is 13.83.
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