homework help 1580

The ##K_”b”## of ##”Z”^-## is ##2.1 × 10^-10##.

The equation for the equilibrium is:

##”Z”^(-) + “H”_2″O” ⇌ “HZ” + “OH”^-##

##[“Z”^-]_0 = “0.350 mol”/”1.0 L” = “0.35 mol/L”##

The ##K_”b”## expression is:

##K_”b” = ([“HZ”][“OH”^-])/([“Z”^-])##

##”[pH](http://socratic.org/chemistry/acids-and-bases/the-ph-concept)” = 8.93##

##”pOH” = 14.00 – 8.93 = 5.07##

##[“OH”^-] = 10^”-pOH” = 10^-5.07 = 8.51 × 10^-6″mol/L”##

##[“HZ”]= 8.51 × 10^-6″mol/L” ##

##[“Z”^-] = (0.35 – 8.51 × 10^-6) ” mol/L” = “0.35 mol/L”##

##K_”b” = ([“HZ”][“OH”^-])/([“Z”^-]) = (8.51 × 10^-6 × 8.51 × 10^-6)/0.35 = 2.1 × 10^-10##

Solution:

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