The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.05 L of KOH w

Question

3 years ago

6

The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.05 L of KOH w

ith 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

Chemistry

2 answers:

Dmitry_Shevchenko [17] · Answer 1 · 2022-05-18T13:57:09+03:00

Dmitry_Shevchenko [17]3 years ago

7 0

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

strojnjashka [21] · Answer 2 · 2022-05-18T12:51:13+03:00

strojnjashka [21]3 years ago

3 0

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