Neutralization Of H2so4 By Koh

The answers for the acid-base reaction between sulfuric acrid and potassium hydroxide are:

a) The balanced neutralization reaction that occurs between H₂Then₄ and KOH is:

H₂So₄(aq) + KOH(aq) ⇄ H₂O(l) + K₂SO₄(aq)

b) The concentration of sulfuric acrid that remains after the neutralization of 0.750 Fifty of 0.480 Thousand H₂So₄ with 0.700 L of 0.290 M KOH is 0.108 Thousand.

a) The acrid-base reaction between H₂Then₄ and KOH is:

H₂And then₄(aq) + KOH(aq) ⇄ H₂O(l) + K₂SO₄(aq)

To residuum the equation, nosotros need to equalize the number of H, Southward, O, and Chiliad atoms in the reactants with the products. On the products side, we accept 2 atoms of K, so nosotros need to add a coefficient of two before KOH:

H₂Then₄(aq) + 2KOH(aq) ⇄ H₂O(l) + K₂And then₄(aq)

Now, nosotros have half-dozen atoms of O on the reactants and 5 on the products side, we need to add together a coefficient of two earlier H₂O:

H₂SO₄(aq) + 2KOH(aq) ⇄ 2H₂O(fifty) + Yard₂Then₄(aq) (ane)

Equation (1) is the balanced reaction.

b) The number of moles of acrid is equal to the number of moles of the base at neutralization. To discover the concentration of the acrid, we need to calculate the number of moles of acid and base:

$n = C*V$

Where:

C: is the molar concentration

Five: is the book

The number of moles of the acid (H₂And so₄):

$n_{a_{i}} = 0.750 L*0.480 mol/L = 0.36 \:moles$

The number of moles of the base (KOH):

$n_{b_{i}} = 0.700 L*0.290 mol/L = 0.203 \:moles$

At the neutralization, we have:

$n_{a} = n_{b}$

$n_{a} = 0.203 \:moles$

Initially, we had 0.36 moles of acid and since 0.203 moles of it reacts with the base, the number of moles of acid that remains after neutralization is:

$n_{f} = 0.36 \: moles - 0.203\: moles = 0.157 \:moles$