Question

What is Ka for HCN(aq) H+(aq)+CN-(aq)?​

Answer 1

c

Step-by-step explanation:

Answer 2

`\tt C.~Ka=([H^+][CN^-])/([HCN])`

Further explanation

According to Arrhenius, acids are substances which, when dissolved in water, release Hions.

An HₓY acid in water will ionize:

HₓY (aq) ⇒ xH⁺ (aq) + Yˣ- (aq)

In general, the weak acid ionization reaction

HA (aq) ⇒ H⁺ (aq) + A⁻ (aq)

Ka's value

`\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}`

So for HCN

HCN(aq)⇒H⁺(aq)+CN⁻(aq)

`\tt Ka=([H^+][CN^-])/([HCN])`