Question

Consider a l buffer made by adding .14mol HCNO and .11 mol KCNO to sufficient water. Ka= 3.5*10^-5 for Cyanic acid HCNO

A) Calcuate the Ph of the buffer before an acid or base is added..... I got 4.35

B) Calculate the pH of the buffer after the addition of .15 mol KOH. Neglect any volume. Please show work My ICE table is wrong i think

C) Starting with the original solution in calculate the ph of the buffer after the addition of .015 mol Hno3. Neglect any volume. Please show work My ICE table is wrong i think as well

Answer 1

Consider a l buffer made by adding .14mol HCNO and .11 mol KCNO to sufficient water. Ka= 3.5*10^-5 for Cyanic acid HCNO

A) Calcuate the Ph of the buffer before an acid or base is added.

We will use Hendersen Hasselbalch equation for calcuating the pH of buffer

pH = pKa + log [salt / acid]

pKa = -logKa = -log (3.5X10^-5) = 4.45

so pH = 4.45 + log 0.11 / 0.14 = 4.345

B) Calculate the pH of the buffer after the addition of .15 mol KOH. Neglect any volume.  

HCNO -->H+ + CN-

KCNO --> K + + CNO-

On addition of KOH, it will react with H+ and will decrese the concentration of acid and will increase the concentrtion of salt by the same amount

Moles of base added = 0.015 ( it can not be 0.15 as it will increase the pH drastically)

so pH = 4.45 + log [0.11 + 0.015 / 0.14 - 0.015]

pH = 4.45 + log [0.125 / 0.125] = 4.45

C) Starting with the original solution in calculate the ph of the buffer after the addition of .015 mol Hno3.

Similarly on addition of acid the concentration of acid will increase and that of salt will decrease

pH = 4.45 + log [0.11 - 0.015 / 0.14 + 0.015]

pH = 4.45 + log [0.0.95 / 0.155] = 4.45 - 0.212 = 4.238