Applications of Acid and Base Equations

Strong acids COMPLETELY dissociate so we can write the reaction: HCl → H+ + Cl-

0.5M

Since the equation is balanced, we can see that there is a 1:1 molar ratio between HCl and H+ so the concentration of H+ will actually be the same!

[H+] =0.5M

plug that into the pH equation: pH=-log[H+]

pH=-log[0.5]

pH=0.30

CH3COOH is not one of the strong acids we had to memorize, so we automatically know that it is a weak acid! Weak acids have INCOMPLETE dissociation, so we will see an equilibrium in the equation:

CH3COOH(aq) + H2O(l) ⇌ CH3COO- + H3O+(aq)

1.0M ? need this to find pH!

Write out an ICE Table:

I 1M / 0 0

C -x +x +x

E 1-x x x

Write out the Ka expression:

$Ka=\frac{\left[CH_3COO^-\right]\left[H_3O^+\right]}{\left[CH_3COOH\right]}$

Plug in the Ka value we were given and plug in the equilibrium concentrations from our ICE table:

$1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[1-x\right]}$

Let's check to see if the denomenator has to stay [1-x] (and we would have to use the quadratic equation) OR we might be able to simplify it to ~1.

When the denominator is in the form of y-x, we can check to see if y/K > 400 and if it is, then we can make the assumption and simplify.

Let's see.... 1/1.8x10-5= 55 555 which is much larger than 400! This means that our denominator was [1-x] but we can simplify it to [1] since y is so large and K is so small so the x won't make a difference.

$1.8x10^{-5}=\frac{x^2}{1}$

Take the square root of both sides:

4.24x10-3=x

If we look back at our ICE table, x=[H+]

so now we can solve for pH:

pH=-log[H+]

pH=-log[4.24x10-3]

pH=2.37 !

Since we know it's a strong base, that means we know it will COMPLETELY dissociate so we can write out the chemical reaction:

Ba(OH)2 → Ba2+ + 2OH- (Always double check your equation is balanced!!)

Here we are given n=0.10 mol, V=1L. We can use the equation n=cv to solve for c:

c=n/v

c=0.10mol/1L

c=0.1M

0.1M 0.2M

(Since Ba(OH)2 has a concentration of 0.1M, OH- must have a concentration of 0.2M since there is a 1:2 molar ratio!)

[OH-]=0.2M

pOH=-log[OH-]

pOH=-log[0.2]

pOH=0.70

We are asked to solve for pH so use the equation: pH + pOH = 14

pH=14-pOH

pH=14-0.70

pH=13.3 (This logically makes sense since we would expect a base to have a very high pH)

NH3 is a weak base so that means it INCOMPLETELY dissociates:

NH3 + H2O ⇌ NH4+ + OH-

5M

Here we will need to make an ICE table, so we can again plug in the equilibrium concentrations into our Ka expression, solve for [OH-] then pOH, then pH !

I 5M / 0 0

C -x +x +x

E 5-x x x

Now write the Kb expression out:

$Kb=\frac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}$

Plug in the Kb value we were given and the equilibrium values from our ICE Table:

$1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[5-x\right]}$

Check to see if we can make the simplification of the denominator from [5-x] to ~5.

y/K >400

5/1.8x10-5=277 778. which is much greater than 400 so we can make the simplifcation!

$1.8x10^{-5}=\frac{\left[x\right]\left[x\right]}{\left[5\right]}$

1.8x10-5(5)=x2

9x10-5=x2

Take the square root of both sides:

9.49x10-3=x

x=[OH-] (see the ICE Table)

Solve for pOH:

pOH=-log[OH-]

pOH=-log[9.49x10-3]

pOH=2.02

The question is asking us to solve for pH so use pH + pOH=14

pH=14 - pOH

pH=14-2.02

pH=11.98 !