Given the following K a \'s: K a (HF) = 7.2x10 -4 K a (HC 2 H 3 O 2 ) = 1.8x10 -

Question

Given the following Ka's:

Ka(HF) = 7.2x10-4

Ka(HC2H3O2) = 1.8x10-5

Ka(HClO) = 3.5x10-8

Ka(HCN) = 4.0x10-10

determine the equilibrium constant for each reaction and indicate whether a single or double arrow would be more appropriate. You may save some time by inputting your answer using the values in exponential format and a divide by sign "/". For example, if calculating the K involves taking 3.5x10-8 and dividing by 4.0x10-10, simply input "3.5e-8/4e-10" into the webassign answer blank. Note: this question seems to be a little sig-fig sensitive, so enter your numeric answers to three significant figures.

Reaction K Arrows HC2H3O2 + CN1- C2H3O21- + HCN _________ ---Select--- single arrow double arrows HClO + CN1- ClO1- + HCN _________ ---Select--- single arrow double arrows HC2H3O2 + ClO1- C2H3O21- + HClO _________ ---Select--- single arrow double arrows HF + CN1- F1-+ HCN _________ ---Select--- single arrow double arrows

Explanation / Answer

I assume that

double arrows should be used on all of them

HC2H3O2 + CN1- -> C2H3O21- + HCN
this is #2 written forward & # 4 in reverse
so its K is [Ka(HC2H3O2) = 1.8x10-5] / [Ka(HCN) = 4.0x10-10]
K = 4.5 X10+4
(a reaction this favorable is often written with 1 arrow --> )
======================================================

HClO + CN1- -> ClO1- + HCN
this is #2 written forward & #1 in reverse
so its K is [Ka(HClO) = 3.5x10-8] / [Ka(HF) = 7.2x10-4]
K = 4.9 X10-5

==========================
HCN + ClO1- -> CN1- + HClO
s is #4 written forward & #3 in reverse
so its K is [Ka(HCN) = 4.0x10-10] / [Ka(HClO) = 3.5x10-8]
K = 5.6 X10-2

========================================================================
HF + CN1- -> F1-+ HCNs is #1 written forward & #4 in reverse
so its K is [Ka(HF) = 7.2x10-4] / [Ka(HCN) = 4.0x10-10]
K = 1.8 X10+6
(a reaction this favorable is often written with 1 arrow --> )

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