Question

Which one of the following solutions will have the greatest concentration of hydroxide ions 0.100 M rubidium 0.100 M magnesium hydroxide 0.100 ammonia 0.100 M beryllium hydroxide 0.100 M hydrochloric acid

Answer 1

Answer : The correct answer is solutions of 0.100 M of Magnesium hydroxide and Beryllium hydroxide.

The Hydroxide ions(OH⁻) are produced when ionic compounds (basic compound ) dissociates in water or compound reacts with water as follows :

MOH (s) + H2O (l) <=> M+ (aq) + OH⁻(aq)

M (aq) + H₂O(l) <=> MH (aq) + OH⁻(aq)

Where MOH and M are compounds.

Concentration of OH⁻ can be found in compound by following steps :

Step 1 : To find acid or base .

First of all we need to find whether the given compound is acid or base by acid - base concept . (image attached )

If a compound is acid , it will not produce OH⁻ ions or negligible amount of OH⁻ ions. So it can be ruled out .

If the compound is base , step 2 should be followed .

Step 2 : To find strength of base .

We need to find whether the given base is strong or weak .

Generally hydroxides of Group 1 and 2 are strong base . Here are some listed strong base ( image attached ) .Base other than these criteria will be weak base .

Step 3 :To find concentration of OH⁻ ions :

If the base is strong , means it will dissociate completely to produce OH⁻ ions . Hence concentration of OH⁻ can be calculated by stoichiometric relation from balanced reaction.

If base is weak , it will not dissociate completely to give all the OH⁻ ions , so we will make ICE table of a balanced reaction for those base and use kb values to calculate concentration of base .

A) Rubidium hydroxide (RbOH ) (0.100 M ) :

According to Arrhenius concept RbOH is a base and from list it is a strong base . Hence it will dissociate completely to give OH⁻ ions .

RbOH + H₂O <=> Rb⁺ + OH⁻

Since the Ratio of RbOH : OH⁻ = 1 : 1 .

$[OH^-] = 0.100 M RbOH * \frac{1 OH^-}{1 RbOH } = 0.100 M OH^-$

[OH⁻] = 0.100 M

B) Magnesium hydroxide , Mg(OH)₂ ( 0.100 M) :

According to Arrhenius concept and since Mg is from group 2 , so Mg(OH)₂ can be said as Strong base . So Mg(OH)₂ will dissociate completely as :

Mg(OH)₂ + H₂O <=> Mg²⁺ + 2 OH⁻

The ratio of OH⁻ : Mg(OH)₂ = 2 : 1

$[OH^- ] = 0.100 M Mg(OH)2 * \frac{2 OH^-}{1 Mg(OH)2}$

[OH⁻ ] = 0.200 M

C) Ammonia (NH3 ) (0.100 M ) :

According to Lewis concept it is base but it neither belongs to list nor group 1 and 2 , Hence it is weak base and will dissociate incompletely as :

NH₃ + H₂O <=> NH₄OH <=> NH₄⁺ + OH⁻

Since it is weak base so ICE table will be formed as:

NH₃ + H₂O <=> NH₄⁺ + OH⁻

Initial(M) 0.100 , - 0 0

Change(M) -x - +x + x

Equilibrium(M) 0.100-x - x x

Now Kb expression will be written as :

$kb = \frac{[NH4^+] * [OH^-] }{[NH3]}$

Kb for NH3 is 1.8 * 10 ⁻⁵ ( from list in image )

Plugging equilibrium values of [NH₄⁺] = [OH⁻] = x and [NH₃] = (0.100-x ) in this expression :

$1.8 * 10 ^-^5 = \frac{x *x}{(0.100 - x )}$

Since kb is small (10⁻⁵ ) , x from (0.100 -x ) can be neglected

$1.8 * 10 ^-^5 = \frac{x *x}{(0.100 )}$

Multiplying both side by 0.100

$1.8 * 10 ^-^5 * 0.100 = \frac{x *x}{(0.100 - x )} * 0.100$

x² = 1.8 * 10⁻⁶

x = 1.34 * 10⁻³

Hence concentration of OH⁻ produced by NH₃ = 1.34 * 10⁻³ M

D) Beryllium hydroxide ,Be(OH)₂ (0.100 M ) :

According to Arrhenius concept it is a base and since it belongs to Group 2 , its is strong base . Hence it will dissociate completely .

Be(OH)₂ + H₂O <=> Be²⁺ + 2 OH⁻

Ratio between OH⁻ : Be(OH)₂ = 2 : 1

$[OH^- ] = 0.100 M Be(OH)2 *\frac{2 OH^-}{1 Be(OH)2}$

Hence [OH⁻] by Be(OH)₂ = 0.200 M

E) Hydrochloric acid , HCl (0.100 M ) :

According to all the concepts HCl is acid . Since it is acid it produce negligible amount of OH⁻ ion . So this option can be neglected .

If all are summarized :

[OH⁻] by RbOH = 0.100 M

[OH⁻] by Mg(OH)₂ = 0.200 M

[OH⁻] by NH₃ = 1.34 * 10⁻³ M

[OH⁻] by Be(OH)₂ = 0.200 M

[OH⁻] by HCl = negligible

Hence Mg(OH)₂ and Be(OH)2 produced highest amount of [OH⁻] .