Question

Molarity for NaOH = 0.0873

2. In a second titration with the same solution of NaOH as used in Question #1, the student weighs out a sample of KHP of 0.359 g. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP. 

3. A monoprotic weak acid with the general formula of HA will react with a base, such as NaOH. Write the neutralization equation which describes the reaction. 

4. If K, for the weak acid, HA is 1.8 x 10, calculate the magnitude of the equilibrium constant for the equation in Question #3. (Note: First write the net ionic equation for the equation in Question #3.) 

5. Another Chem 1515 student was given a 50.0 mL sample of the weak acid in Problem #4. The concentration of the weak acid is given as 0.15 M. Using a pipette, she delivers 25.0 mL of the weak acid into a 100 mL beaker. She then titrated this solution using a standardized solution of 0.21 M NaOH. 

• a) How many moles of the weak acid were added to the beaker? 

• b) How many moles of NaOH are required to neutralize (reach the equivalence point) the sample of weak acid? 

Answer 1

2. mass of KHP used = 0.359 g

molar mass of KHP = 204 g/mol

so, moles of KHP = 0.359 g / 204 g/mol = 1.76*10^-3 moles

NaOH = 0.0873 M = 0.0873 mol /L

Since in reaction, 1 mol of KHP = 1 mol of NaOH (1 L = 1000 ml)

So, volume of NaOH required = 1.76*10^-3 moles / (0.0873 mol /L ) = 0.0202 L = 20.2 ml

3. Neutralization reaction with monoprotic base :

HA (aq) + NaOH (aq) $\rightleftharpoons$ NaA (aq) + H₂O (l)

or HA (aq) + Na⁺ (aq) + OH^- (aq) $\rightleftharpoons$ Na⁺ (aq) + A^-(aq) + H₂O (l)

4. Net ionic equation :

HA (aq) + OH^- (aq) $\rightleftharpoons$ A^-(aq) + H₂O (l)

Keq = (A^-(aq)) (H₂O (l)) / ( HA (aq) ) (OH^- (aq))

since water in excess, we can write : Keq = (A^-(aq)) / ( HA (aq) ) (OH^- (aq))

we can write ,  Keq = (A^-(aq))(H₃O⁺ (aq)) / ( HA (aq) ) (OH^- (aq))(H₃O⁺ (aq)) ...................1

we have,

HA (aq) + H₂O $\rightleftharpoons$ A^-(aq) + H₃O⁺ (aq)  

since water in excess, we can write

Ka = (A^-(aq)) (H₃O⁺ (aq)) / ( HA (aq) ) = 1.8*10^-5 .......................2

from 1 and 2,

we have, Keq = Ka / (OH^- (aq))(H₃O⁺ (aq)) = Ka /Kw (Kw = (OH^- (aq))(H₃O⁺ (aq) = 10^-14)

^So, Keq = 1.8 * 10⁹

^5. (a) 25.0 ml of weak acid of 0.15 M added to beaker :

moles of weak acid added = 0.15 mol/L * 25.0 ml / 1000 ml /L = 3.75*10^-3 moles

(b)

we have,   HA (aq) + OH^- (aq) $\rightleftharpoons$ A^-(aq) + H₂O (l)

Since in reaction, 1 mol of HA = 1 mol of NaOH

so moles of NaOH required = 3.75*10^-3 moles