a) 0.21 M HCl
ph=
b)2.8×10^−2 M HNO3.
ph=
c)a solution that is 5.3×10^−2 M in HBr and 2.0×10−2 M in HNO3.
ph=
d)a solution that is 0.680%HNO3 by mass (Assume a density of 1.01g/ml for the solution.)
ph=
can you please explain in detail how you got the answers? Thanks
3 Answers

a) 0.21 M HCl
[H+] = 0.21M
pH = log(0.21) = 0.678
pOH = 140.678 = 13.322
[OH] = 10^13.22 = 4.76 x 10^14M
b)2.8×10^−2 M HNO3.
[H+] = 2.8E2
pH = log(2.8E2) = 1.55
pOH = 141.55 = 12.44
[OH] = 10^12.44 = 3.57 x 10^13M
c)a solution that is 5.3×10^−2 M in HBr and 2.0×10−2 M in HNO3.
[H+] = 5.3E2 + 2.0E2 = 7.3E2
pH = log(7.3E2) = 1.14
pOH = 141.14 = 12.86
[OH] = 10^12.86 = 1.38 x 10^13M
d)a solution that is 0.680%HNO3 by mass (Assume a density of 1.01g/ml for the solution.)
1L of solution has a mass of 1010g.
The mass of HNO3 in 1010g of solution = 1010g x .00680 = 6.87gHNO3
6.87g x [1mole / 63g ] = 0.109moles in 1L = 0.109M
[H+] = 0.109M
pH = log(0.109) = 0.963
pOH = 14.963 = 13.037
[OH] = 10^13.037 = 9.17 x 10^14M
Hope this helps.

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