Question

Find the pH of the following solutions: a mixture of 10.0 mL NaOH solution having pH 11.00 and 10.0 mL HClO4 having pH 1.00.

2.) Using activities calcalute the pH and concentration of H+ in pure water containing 0.05 M CaCl2 at 25 degrees C

Answer 1

1) 10.0 mL NaOH solution having pH 11.00 and 10.0 mL HClO4 having pH 1.00.

Strong acid and strong base:

HClO4: [H⁺] = 10^-pH = 10^-1 = 0.1 mol/L

NaOH: [OH^-] = 10^-14+pH = 10^-14+11 = 0.001 mol/L

After mixing following reaction occurs:

H⁺ + OH^- = H₂O

New concentration of H⁺ will be:

[H⁺] = (0.01 L * 0.1 mol/l - 0.01 L* 0.001 mol/L (consumed during neutralization)) / (0.01 L + 0.01 L) = 0.0495 mol/L

pH = -log[H⁺] = -log 0.0495 M = -(-1.31) = 1.31

2) Using activities calcalute the pH and concentration of H+ in pure water containing 0.05 M CaCl2 at 25 degrees C.

Water ionization:

H₂O = H⁺ + OH^-

Salt of strong base+strong acid = neutral!

[H+] = [OH-] = (Kw)^1/2 = (10^-14)^1/2 = 10^-7 mol/L

Ionic strenght of solution:

I = 1/2*Sum(c_iz_i²)

c-iones concentration

z-iones charge

CaCl₂ = Ca²⁺ + 2Cl^-

[Ca²⁺]=1/2[Cl^-]=c(CaCl₂)=0.05 mol/L

I = 1/2(4*0.05 mol/L + 2*1*0.05 mol/L) = 1.5 mol/L.

For activity:

a(H⁺) = y*[H⁺]

y-activity coeficient

log(y)=0.5*z²*(I)^1/2 / (1+(I)^1/2)

Using above formula for monovalent ion:

y=0.725

a(H⁺) = 0.725*[H⁺]=7.25*10^-8

pH = -loga(H⁺) = -log 7.25*10^-8 = -(-7.14) = 7.14