Question

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).

Answer 1

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

Answer 2

Answer:

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