A. Write an equation for determining the number of moles of NaOH that are added to the flask based on [NaOH] and volume of NaOH titrant (mL NaOH):

B. Write a similar expression for the number of moles of H2SO4 in the flask based on [H2SO4] and the volume of H2SO4 (mL).

Substitute your expressions from A and B into this equation and solve for [H2SO4]

From the equation for molarity (number of moles of a substance dissolved in one liter of solution), we can get an equation for determining the number of moles:

c=n/V ⇒ n=c×V

Now, we can write required equations:

A n(NaOH)=c(NaOH)×V(NaOH)

B n(H₂SO₄)=c(H₂SO₄)×V(H₂SO₄)

From the reaction, we can conclude that NaOH and H₂SO₄ have following stoichiometric ratio:

2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂0

n(NaOH) : n(H₂SO₄) = 2 : 1

So, we can write:

n(NaOH) = 2 × n(H₂SO₄)

And if we connect the above equations, we get the equation for solving c(H2SO4):

c(NaOH)×V(NaOH) = 2×c(H₂SO₄)×V(H₂SO₄)

c(H₂SO₄) = (c(NaOH)×V(NaOH)) / (2×V(H₂SO₄))

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-02T15:49:49+01:00

The concentration of [H2SO4] is obtained from; [NaOH] × VNaOH/ 2× VH2SO4

The equation of the reaction is;

2NaOH(aq) + H2SO4(aq) ------> Na2SO4(aq) + 2H2O(l)

Number of moles of H2SO4(nH2SO4) = [H2SO4] × VH2SO4

Number of moles of NaOH(nNaOH) = [NaOH] × VNaOH

Where;

[H2SO4] = concentration of H2SO4

VH2SO4 = volume of H2SO4

[NaOH] = concentration of NaOH

VNaOH = volume of NaOH

From the balanced reaction equation;

2 moles of NaOH reacts with 1 mole of H2SO4

2nH2SO4 = nNaOH

2[H2SO4] × VH2SO4 = [NaOH] × VNaOH

[H2SO4] = [NaOH] × VNaOH/ 2× VH2SO4

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