Question

1 If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl? (5 points)

2 If it takes 27.4 mL of 0.050 M H2SO4 to neutralize 357 mL of KOH solution, what is the concentration of the NaOH solution? (5 points)




3 If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H2SO4), what is the concentration of the H2SO4 solution? (5 points)


4 Explain the difference between an endpoint and equivalence point in a titration. (5 points)

Answer 1

Answer:- (1) 0.075 M, (2) 0.00768M and (3) 0.106M.

Solution:- (1) Molarity is moles of solute per liter of solution. From given volume and molarity we can calculate the moles of NaOH.

We convert mL to L and multiply by molarity to get the moles.

`67mL*((1L)/(1000mL))((0.15mol)/(1L))`

= 0.01005 mol NaOH

NaOH and HCl react in 1:1 mol ratio which is clear from the below equation:-

`NaOH(aq)+HCl(a)\rightleftharpoons NaCl(aq)+H_2O(l)`

As they react in 1:1 mol ratio, the moles of HCl will also be 0.01005. Volume of HCl is given. We convert the mL to L and divide the moles by it to get the molarity of HCl.

`(0.01005mol)/(134mL)((1000mL)/(1L))`

= 0.075 M

So, the molarity of HCl solution is 0.075M.

(2) This one is also similar to the previous one. Only the mol ratio will be different as per the equation:

`2KOH(aq)+H_2SO_4(aq)\rightleftharpoons K_2SO_4(a)+2H_2O(l)`

`((27.4mLH_2SO_4)/(1))((1L)/(1000mL))((0.050mol)/(1L))((2molKOH)/(1molH_2SO_4))((1)/(357mL))((1000mL)/(1L))`

= 0.00768M KOH

So, the concentration of KOH(or if it is even NaOH) is 0.00768 M.

(3) This one is exactly similar to the second one. The balanced equation is:

`2NaOH(aq)+H_2SO_4(aq)\rightleftharpoons Na_2SO_4(a)+2H_2O(l)`

`((55mL)/(1))((1L)/(1000mL))((0.5molNaOH)/(1L))((1molH_2SO_4)/(2molNaOH))((1)/(130mL))((1000mL)/(1L))`

= 0.106M
`H_2SO_4`

So, the molarity of the sulfuric acid is 0.106 M.

(4) In acid base titration, the end point is the point at which the indicator changes into color where as the equivalence point is the point at which the moles of acid and base are in their stoichiometric ratio.

For example if we have 0.2 moles of sulfuric acid then there must be 0.4 moles of NaOH at equivalence point since they react in 1:2 mol ratio.