calculate the ph, ph=-log(AH+) of a 0.0200 M solution of Ca(NO3)2 to which 5.87 x10^-7 moles...
calculate the ph, ph=-log(AH+) of a 0.0200 M solution of Ca(NO3)2 to which 5.87 x10^-7 moles of HCL have been added per L of solution. Include ionic strength effects in this calculation and equilibrium constant.
Homework Answers
we have
0.0200 M solution of Ca(NO3)2.
Ca(NO3)2 ........> Ca^2+ + 2 NO3-
[Ca^2+] = 0.02 M
and
[NO3-] = 2 * 0.02 = 0.04 M
ionic strength (v) = 1/2 * [0.02 * (+2)^2 + 0.04 * (-1)^2] = 0.06 M
we know,
log (activity coefficient) = - A * Z+ * Z- * sqrt (v)
or
log (activity coefficient) = - 0.51 * 2 * 1 * sqrt (0.06)
or
activity coefficient = 0.5625
thus
HCl is strong acid.
activity of H+ = 5.87 * 10^-7 * 0.5625 = 3.302 * 10^-7 M
pH = -log (activity of H+) = - log (3.302 * 10^-7 ) = 6.48 (answer)
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calculate the ph, ph=-log(AH+) of a 0.0200 M solution of
Ca(NO3)2 to which 5.87 x10^-7 moles...
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