If 1.0 L of 0.2 M Ca(OH)2 is added to 2.5 L of 0.1 M HCl, the pH of the final solution is?
can someone show me the working out of this question? im very confused by it. thankyou
moles Ca(OH)2 = M x L = 0.2 x 1.0 = 0.2 mole.
moles HCl = M x L = 0.1 x 2.5 = 0.25 mole.
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
This is a limiting reagent problem and you need to determine which is the limiting reagent. The pH will be determined by the OTHER reagent.
Convert moles Ca(OH)2 to moles CaCl2.
Convert moles HCl to moles CaCl2.
The two answers will be different (and only one can be correct). The SMALLER one is ALWAYS the correct value in limiting reagent problems and the reagent producing the smaller value will be the limiting reagent.
Now convert the limiting reagent moles to moles of the non-limiting reagent, subtract from the original value of moles, and the excess will determine the (H^+) or (OH)2 (as the case may be) and pH = -log(H^+). Post your work if you get stuck. The above sounds like it can be confusing; however, if you follow step by step things work out very well.
To find the pH of the final solution after mixing Ca(OH)2 and HCl, we need to determine if a reaction occurs between the two substances and if so, which products are formed.
First, let's write down the balanced chemical equation for the reaction between Ca(OH)2 and HCl:
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
This equation shows that one molecule of Ca(OH)2 reacts with two molecules of HCl to produce one molecule of CaCl2 and two molecules of water.
Now, let's calculate the number of moles of Ca(OH)2 and HCl in the given solutions:
For Ca(OH)2:
0.2 M = 0.2 moles of Ca(OH)2 per liter
Volume = 1.0 L
Number of moles of Ca(OH)2 = 0.2 moles/L × 1.0 L = 0.2 moles
For HCl:
0.1 M = 0.1 moles of HCl per liter
Volume = 2.5 L
Number of moles of HCl = 0.1 moles/L × 2.5 L = 0.25 moles
Next, let's determine the limiting reactant. The limiting reactant is the one that will be completely consumed in the reaction. We can see from the balanced equation that the reaction ratio between Ca(OH)2 and HCl is 1:2. Therefore, if we have less HCl than required for this ratio, HCl will be the limiting reactant.
In this case, we have 0.25 moles of HCl and 0.20 moles of Ca(OH)2. Since we need twice the moles of HCl compared to Ca(OH)2, the HCl will be completely consumed, leaving some Ca(OH)2 unreacted.
The excess Ca(OH)2 will remain in the solution and react partially with water to form OH- ions. Since OH- ions are the conjugate base of water, they will increase the pH of the solution.
To calculate the concentration of OH- ions, we need to find the moles of OH- ions formed from the remaining Ca(OH)2. Since the reaction ratio between Ca(OH)2 and OH- is 1:2, the moles of OH- ions formed will be twice the moles of remaining Ca(OH)2.
The moles of remaining Ca(OH)2 = 0.20 moles - 0.25 moles = -0.05 moles (negative value means it is completely consumed)
Moles of OH- ions formed = 2 × (-0.05 moles) = -0.10 moles
Since we started with 2.5 L of solution, the concentration of OH- ions can be calculated as follows:
Concentration = Moles / Volume
Concentration = (-0.10 moles) / 2.5 L = -0.04 M (negative value still represents the concentration)
Finally, let's calculate the pOH and pH of the final solution:
pOH = -log10(0.04) = 1.4 (rounded to one decimal place)
pH = 14 - pOH = 14 - 1.4 = 12.6
Therefore, the pH of the final solution is 12.6.