Calculate (Cr3+) in equilibrium with Cr(OH)4- when 0.010mol of Cr(NO3)3 is disso

Question

Calculate (Cr3+) in equilibrium with Cr(OH)4- when 0.010mol of Cr(NO3)3 is dissolved in 1L of solution buffered at pH 10.0.

Can you please show an ice chart and explain how to solve this?

Explanation / Answer

Cr3+(Aq)+4OH-(a)----->Cr(OH)4- Kf=8x10^29 POH=14.0-10.0=4.0 |OH|=antilogPOH=1.0X10^ -4 Due to lage Kf all of Cr3+ is converted into the complex ion and some subsiquently dissociates back to Cr3+,then at equillibrrium, Cr3+(Aq) + 4OH-(a) ----->Cr(OH)4- x 1.0X10^ -4 0.010-x Kf=|Cr(OH)4-| / |Cr3+| |OH-|4=0.010-x / x(1.0X10^ -4 )=8x10^29 Cr3+=x=1.2x10^ -16 M

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