An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of t
Answer:
When we add 0.33L of KOH only the HCl solution will be neutralized. When we add <u>1.4 L</u> of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.
Explanation:
<u>Step 1:</u> Data given
The solution has 0.100 M HCl and 0.210 M H2SO4
Molarity KOH = 0.150 M
Volume of acid solution = 500 mL = 0.5 L
<u>Step 2: </u>Calculate moles of HCl
Moles HCl = Molarity HCl * volume
Moles HCl = 0.100 M * 0.5 L
Moles HCl = 0.05 moles
<u>Step 3:</u> Calculate moles of H2SO4
Moles H2SO4 = 0.210 M * 0.5 L
Moles H2SO4 = 0.105 moles
<u>Step 4:</u> The balanced neutralization reaction of KOH with HCl can be written as:
KOH + HCl → KCl + H2O
The mole ratio is 1:1
This means to neutralize 0.05 moles HCl, we need 0.05 moles KOH
<u>Step 5:</u> The balanced neutralization reaction of KOH with H2SO4 can be written as:
2KOH + H2SO4 → K2SO4 + 2H2O
The mole ratio KOH: H2SO4 is 2:1
This means to neutralize 0.105 moles H2SO4 we need 0.210 moles KOH
<u>Step 6: </u>Calculate volume of KOH needed to neutralize the solution
To neutralize the HCl solution: 0.05 moles / 0.150 M = 0.33 L KOH needed
To neutralize the H2SO4 solution: 0.210 moles / 0.150 M = 1.4 L KOH needed
When we add 0.33L of KOH the HCl solution will be neutralized. When we add 1.4 L of KOH the HCl and H2SO4 solution will be neutralized.
Explanation:
Mass = volume × density
Mass = 652 cm³ × 21.45 g/cm³
= 13985.4 g
Explanation:
Answer:
Precipitation reaction
Explanation:
Given that solution A was mixed with solution B, the solution turned cloudy. The test is not warm and no bubbles visible. This means that the precipiate is formed.
The concept is when two colourless solutions react to form a cloudy precipitate that settles at bottom of a solution then the reaction is said to be a precipitation reaction.
An example can be the Reaction of Silver nitrate with common salt.