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# Answer to Question #163989 in General Chemistry for Keturah Millington

Question #163989

40 cm3of a 0.2 moldm-3solution of H2SO4 reacts completely with 25 cm3 of a NaOH solution to produce sodium sulphate and water.

Write a balanced equation for this reaction.

Write an ionic equation for the reaction.

Determine the number of moles present in the 40cm3 of the acid.

What is the ratio in which the acid and alkali reacts?

What is the number of moles present in the 25cm3 of alkali?

How many moles will be present in 800cm3 of the alkali.?

What is the concentration of the alkali in moldm-3

What is the concentration of the alkali in gdm-3

Volume of solutions/cm3 Titration number

1 2 3 4

Final burette readings 37.50 38.20 40.40

1
2021-02-16T07:48:28-0500

Balanced Chemical Equation is given by-

"2NaOH+H \n_2\n\u200b\t\n SO \n_4\n\u200b\t\n \u2192Na \n_2\n\u200b\t\n SO \n_4\n\u200b\t\n +2H \n_2\n\u200b\t\n O+Heat"

Ionic Equation of reaction is-

"2H^+ + (SO_4)^{2-} + 2Na^+ + 2(OH)^- \u21922Na^+ + (SO_4)^{2-} + 2H_2O"

Number of moles present in "40cm^3" of acid is-

=Molarity "\\times \\text{concentration}"

="\\dfrac{0.2}{1000}\\times 40=0.008" mol

Ratio in which acid and alkali react is "1:2"

From The above reaction

2 moles of "H_2SO_4" react with "1" mole of NaoH

Moles of NaoH ="2\\times" 0.008=0.016

Molarity of Alkali= "\\dfrac{\\text{No.of moles }}{Volume}"

= "\\dfrac{0.016\\times 1000}{25}=0.64\\text{mol}" dm"^{-3}"

Number of moles in presen in "800cm^3" of Alkali

="\\dfrac{0.64\\times 800}{1000}=0.512" mol

Concentration of Alkali="0.64\\text{mol dm}^{-3}"

Concentartion of Alkali in "\\text{gdm}^{-3}"

="\\dfrac{0.016\\times 40\\times 1000}{25}=25.6\\text{gdm}^{-3}"

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