Given a balanced chemical equation between H2SO4 (aq) and KOH (aq) H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l) What volume (in mL) of 0.43 M H2SO4 (aq) solution is necessary to completely react with 120 mL of 0.35 M KOH (aq) ? Note: (1) The unit of volume of H2SO4 (aq) is in mL (2) Insert only the numerical value (integer) of your answer (do not include the units or chemical in your answer).

48.84mL

Explanation:

H2SO4 + 2KOH → K2SO4 + 2H2O

From the question:

nA = 1

nB = 2

From the question given we obtained the following information:

Ma = 0.43M

Va = ?

Mb = 0.35M

Vb = 120mL

Using MaVa / MbVb = nA/nB, we can easily find the volume of the acid required. This is illustrated below:

MaVa / MbVb = nA/nB

0.43 x Va / 0.35 x 120 = 1/2

Cross multiply to express in linear form

2 x 0.43 x Va = 0.35 x 120

Divide both side by the (2 x 0.43)

Va = (0.35 x 120) / (2 x 0.43)

Va = 48.84mL

Therefore, the volume of H2SO4 required is 48.84mL

The volume of the H2SO4 solution is 48.8 mL

Explanation:

Step 1: Data given

Molarity of H2SO4 solution = 0.43 M

Volume of KOH = 120 mL

Molarity KOH = 0.35 M

Step 2: The balanced equation

H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l)

Step 3: Calculate the volume of H2SO4

b*Ca*Va = a*Cb*Vb

⇒ b = the coefficient of KOH = 2

⇒ Ca = the molarity of H2SO4 = 0.43 M

⇒ Va = the volume of H2SO4 = ?

⇒ a = the coefficient of H2SO4 = 1

⇒ Cb = the molarity of KOH = 0.35 M

⇒ Vb = the volume of KOH = 0.120 L

2*0.43 * Va = 1*0.35 * 0.120 L

0.86 Va = 0.042

Va = 0.0488 L = 48.8 mL

The volume of the H2SO4 solution is 48.8 mL