ka=[H+][CN-]/[HCN]
The net ionic equation for the reaction is: CN^- + H^+ -> HCn
NH4OH + HC2H3O2 ---> NH4C2H3O2 + H2ONH4+ + OH- + H+ + C2H3O2- ---> NH4+ + C2H3O2- + H2OOH- (aq) + H+ (aq)---> H2O (l)
The conjugate base of HCN is CN-. It is formed when HCN donates a proton (H+) and becomes negatively charged.
HCN - Hydrogen cyanide The conjugate acid of CN- is HCN. HCN stands for hydrogen cyanide. The conjugate acids are a combination of a strong acid and a low base.
Ka= [H+] [H2BO3-] / [h3BO3] (Apex)
HCN(aq) ==> H^+(aq) + CN^-(aq)Ka = [H+][CN-]/[HCN] and the value can be looked up in a book or on line.
The net ionic equation for the reaction is: CN^- + H^+ -> HCn
A pH of 4.29 corresponds to a hydrogen ion concentration of 10-4.29, which equals 10-5 X 10+0.71, since 0.71 - 5 = -4.29. 10+0.71 = 5.1, to the justified number of significant digits. By definition, Ka = [H+]X[CN-]/[HCN], and in the absence of other sources of cyanide anions, [H+] = [CN-]. Within the number of significant digits given, [HCN] = 0.16, and Ka then = (5.1 X 10-5)2/0.16 = 1.6 X 10-8.
NH4OH + HC2H3O2 ---> NH4C2H3O2 + H2ONH4+ + OH- + H+ + C2H3O2- ---> NH4+ + C2H3O2- + H2OOH- (aq) + H+ (aq)---> H2O (l)
H+ + cn- ---> hcn
The conjugate base of HCN is CN-. It is formed when HCN donates a proton (H+) and becomes negatively charged.
HCN - Hydrogen cyanide The conjugate acid of CN- is HCN. HCN stands for hydrogen cyanide. The conjugate acids are a combination of a strong acid and a low base.
ka=[H+][NO2_]/[HNO2]
Ka= [H+] [H2BO3-] / [h3BO3] (Apex)
Ka= [h+][HCO3-]/[H2CO3]
C2H3O2-(aq) H+(aq) H3O+(aq) OH-(aq) HCN(aq)Those are the choices that could be given, but the answer is HCN. It can't be H+ or OH- because the water is de-ionized and H3O+ is not possible.
not sure