The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.05 L of KOH with unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.10 M H2SO4 is dripped into the KOH solution. After exactly 0.033 L of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH.

H2SO4+2KOH--->K2SO4+2H2O

Question

Chemistry

Sylvia

19 August, 10:18

The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.05 L of KOH with unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.10 M H2SO4 is dripped into the KOH solution. After exactly 0.033 L of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH.

H2SO4+2KOH--->K2SO4+2H2O

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Answers (2)

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Eduardo · Answer 1

From the chemical equation given:

H2SO4+2KOH--->K2SO4+2H2O

the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.

No. of moles of KOH = 2 * no. of moles of H2SO4

=2*0.1*0.033

The concentration of KOH = no. of moles / volume

=2*0.1*0.033/0.05

=0.132M

Aidyn Zuniga · Answer 2

chem eqn is H2SO4+2KOH--->K2SO4+2H2O

1x H2SO4 reacts with 2x KOH

0.033L of 0.10M H2SO4 has 0.033 x 0.10 = 0.0033 moles

2x 0.0033 = 0.0066 moles of KOH

conc of KOH = moles/vol = 0.0066/0.05

=0.132M