Constants | Periodic Table Part A A titra concentration of a solution by allowin

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Constants | Periodic Table Part A A titra concentration of a solution by allowing it to react with another solution of known concentration (called a standard solution). Acid-base reactions and oxidation- reduction reactions are used in titrations. For example to find the concentration of an HCl solution (an acid), a standard solution of NaOH (a base) is added to a measured volume of HCl from a calibrated tube called a buret. An indicator is also present and it will change color when all the acid has reacted. Using the concentration of the standard solution and the volume dispensed, we can calculate molarity of the HCl solution tion is a procedure f for determining the A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2S04) What was the molarity of the KOH solution if 18.7 mL of 1.50 M H2 SO4 was needed? The equation is 2KOH(aq) H2SO4(aq)-K2SO4 (aq) 2H2Ol) Express your answer with the appropriate units. View Available Hint(s) molarityValue Units Submit Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H202, can be titrated against a solution of potassium permanganate, KMnO. The following equation represents the reaction 2KMno,(a)+HO(a)+3H,So,(a)+ 302 (g) +2MnSO(aq+KS04aq) 4H20(1)

Explanation / Answer

H2SO4 and KOH are strong acid and base respectively.

Volume of KOH solution(V1) = 90 ml

Volume H2SO4 of required to neutralise it (V2)=18.7 ml

Strength of H2SO4 (S2) =1.50 M

Let strength of KOH solution be S1.

We know V1 x S1= V2 x S2

i.e. 90 ml x S2 = 18.7 ml x 1.50 M

Therefore, S2= 0.31 M

So, molarity of KOH solution = 0.31 M

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