6. What are the equilibrium concentrations of a 0.80 M HCN solution for each species (in M to two decimal places)? HCN(aq) = H(aq) + CN (aq) K = 4.9 x 10 -10
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![6. What are the equilibrium concentrations of a 0.80 M HCN solution for each species (in M to
two
decimal places)?
HCN(aq) = H(aq) + CN¯(aq)
Kc = 4.9 × 10-10
A: [HCN] = 0.80, [H] = 1.98×105, [CN] = 1.98×10
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