Try a few problems - these are like the ones we will do in the lab. 1. A solutio
ID: 977119 • Letter: T
Question
Try a few problems - these are like the ones we will do in the lab. 1. A solution of NaOH of unknown concentration reacts with a standard solution of HC1. In the reaction 14.0mL of 0.500M HC1 are required to neutralize 10.0mLof the unknown NaOH solution. What is the concentration of the NaOH solution? 2. 30.0mL of KOH, of unknown concentration, reacts with 37.4mL of 0.301M HC1. What is the concentration of the KOH? 19.4 mL of 0.250M H2S04 reacts with 24.8mL of NaOH solution. What is the concentration of the NaOH, be careful the acid is diprotic. 10.6 Buffers A buffer solution maintains the pH of a solution by neutralizing added acids and bases. In a buffer solution an acid must be present to react with any OH- ions that are added, and a base must be available to react with any H3O+ ions that are added. Usually a combination of an acid-base conjugate pair is used in buffers. A weak acid and a salt containing its conjugate base or a weak base and a salt of the base containing the conjugate acid.Explanation / Answer
Ans 1.
Concentration of NaOH solution can be calculated by using following equation:
Equivalence of NaOH = equivalence of HCl
M1V1 = M2V2
Where,
M1 = concentration of NaOH,
V1 = volume of NaOH
M2 = concentration of HCl
V2 = concentration of HCl
Therefore, M1 = M2V2 / V1
M1 = 14.0mL x 0.500M / 10.0mL
M1 = 0.700M
Thus, concentration of the NaOH solution is 0.700M
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Ans 2.
Concentration of KOH can be calculated by using following equation,
Equivalence of KOH = equivalence of HCl
M1V1 = M2V2
Where,
M1 = concentration of KOH
V1 = volume of KOH
M2 = concentration of HCl
V2 = volume of HCl
Thus,
M1 = M2V2/ V1
M1 = 0.301M x 37.4mL / 30.0 mL
M1 = 0.375 M
Therefore, concentration of KOH is 0.375 M
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Ans :
For diprotic acid H2SO4 and NaOH reaction following equation can be written,
Equivalence of H2SO4 = equivalence of NaOH
2 x M1V1 = M2V2
Where,
M1 = concentration of H2SO4
V1 = volume of H2SO4
M2 = concentration of NaOH
V2 = volume of NaOH
Thus,
M2 =2 x M1V1 / V2
M2 = 2 x 0.250M x 19.4mL / 24.8mL
M2 = 0.391M
Therefore, concentration of NaOH is 0.391M