Given:C=0.45M We know that the dissociation constant (Ka) of HCN=6.28 X10-10 Solution: KCN is salt of strong acid and weak base. Kh=Kw/Ka Kh=1 X 10-14/6.2 * 10-10 Kh =1.613 X10-5 ---- Kh=h2C h2=Kh/C =1.613 X 10-5/0.45 =0.3584 X10-4 ---- h=\/0.3584 X 10-4 h =0.6 X 10-2 ---- For a salt of weak acid and strong base, [OH-] = h X C =0.6 * 10-2 X 0.45 [OH-] =2.7 X 10-3 ---- pOH = -log[OH-] =-[log (2.7 X 10-3)] =-[-3 log 10 + log 2.7] =-[-3 + 0.4314] =3 - 0.4314 pOH = 2.5686 ---- pH + pOH = 14 pH = 14 - pOH =14 - 2.5686 pH = 11.4314 ---- Therefore the pH of 0.45M KCN is = 11.4314.
Hydrolysis of KCN yields HCN + KOH so the solution will have a pH >7.KCN + H2O ==> KOH + HCN
CN^- + H2O ==> HCN + OH-
Kb = 1x10^-14/6.2x10^-10 = 1.61x10^-5
1.61x10^-5 = [HCN][OH-]/[KCN] = (x)(x)/0.0765
x^2 = 1.2x10^-6
x = 1.1x10^-3 = [OH-]
pOH = 2.96
pH = 14 - 2.96
pH = 11
KCN is a salt of strong base weak acid. Therefore its water solution will be basic. Hence, its pH will be greater than 7. ^So what is its pH?
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
CuSO4 + KCN --->Cu(CN)2 + K2SO4
-log(0.1 M) = 1 pH
- log(0.001 M NH4Cl) = 3 pH =====
Yes. Since KCN is a salt of strong base and weak acid,its water solution will be basic.Therefore,0.1M KCN have pH greater than 7,its pH is 11.
KCN is a salt of strong base weak acid. Therefore its water solution will be basic. Hence, its pH will be greater than 7. ^So what is its pH?
Potassium iodide is neutral. So any solution of potassium iodide would have a pH of 7. One note: no substance has a particular pH as pH depends on both the acidity/basicity of a substance and how concentrated it is.
The answer will depend on the quantity of KCN.
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
Since potassium cyanide (KCN) is highly poisonous, the addition of KCN to water might be caused by an intention to kill.
-log(0.1 M) = 1 pH
KCN kills a human being in 0.5 second.
CuSO4 + KCN --->Cu(CN)2 + K2SO4
To answer this you need a roman numeral on gold to know the charge on it. Assuming it would be (I)... the formula would be KAu(CN)2
- log(0.001 M NH4Cl) = 3 pH =====
M1 and v1 --> KCN M2 and v2 --> HClO ------SO, M1 * v1=M2 * v2 0.36 * 0.037L=M2 * 0.075L 0.013=M2 * 0.075 M2=0.013/0.075 M2=0.17 M