You have a solution of .5 L of 1.74g of NH3 and 4.15g of (NH4)2SO4. How many ml of 14M HCl is required for a pH of 9.00?
I don't know how to go finish this one. I thought about figuring the concentrations of each (NH3, NH4), making an ice table and solving for the additino of H+. But then i don't know how to get ml, cuz what i solved would be in conc of H+ wouldn't it? :s
(Explanation would be greatful, thanks)
Sorry for the double post, but i want to make sure you see it :$
I worked that for you, in detail, below at your earlier post. It took quite a bit of typing which is why I was so slow.
http://www.jiskha.com/display.cgi?id=1321314433
Praise the lord. Thank you, i am forever greatful. Kinda pissed that i had the answer alll along and just thought it was in M, not moles. Otherwise i would have known to divide it. But owells, i can't thank you enough man, you're too good :DDDDDDDDDD
To find out how many milliliters of 14M HCl is required for a specific pH, you will need to consider the chemical reactions involved and use the concept of stoichiometry. Here's how you can approach this problem:
First, let's write down the chemical reactions that will occur in the solution:
1) NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
2) HCl (aq) → H+ (aq) + Cl- (aq)
The addition of HCl will increase the concentration of H+ ions, which will affect the pH of the solution. The equilibrium reaction between NH3 and H2O will also play a role in determining the pH.
Let's break down the solution and calculate the concentrations of NH3 and NH4+ ions:
NH3:
- Mass of NH3 = 1.74g
- Molar mass of NH3 = 17.03 g/mol
- Moles of NH3 = mass/molar mass = 1.74g / 17.03 g/mol = 0.102 moles
NH4+:
- Mass of (NH4)2SO4 = 4.15g
- Molar mass of (NH4)2SO4 = 132.14 g/mol
- Moles of (NH4)2SO4 = mass/molar mass = 4.15g / 132.14 g/mol = 0.031 moles
We know that 1 mole of NH3 reacts with 1 mole of HCl according to the balanced equation.
From the balanced equation of the reaction between NH3 and HCl, we can conclude that the amount of HCl required to react with the NH3 is also 0.102 moles.
Now, let's use the concentration of HCl to determine the volume required for a pH of 9.00:
Given:
- HCl concentration = 14 M
- Desired pH = 9.00
pH is defined as the negative logarithm of the concentration of H+ ions. In this case, we can use the equation:
pH = -log[H+]
To find the concentration of H+ ions required for a pH of 9.00, rearrange the equation:
[H+] = 10^(-pH)
[H+] = 10^(-9.00)
[H+] = 1.00 × 10^(-9) M
This means that the concentration of H+ ions required for a pH of 9.00 is 1.00 × 10^(-9) M.
Since the reaction between NH3 and HCl is 1:1, we need the same concentration of HCl as the concentration of NH3. Therefore, the volume of HCl required can be calculated using the formula:
Volume of HCl (in L) = Moles of HCl / Concentration of HCl
Given:
- Moles of HCl = 0.102 moles
- Concentration of HCl = 14 M
Volume of HCl (in L) = 0.102 moles / 14 M = 0.0073 L
To convert this to milliliters (ml):
Volume of HCl (in mL) = 0.0073 L x 1000 mL/L = 7.3 mL
Therefore, approximately 7.3 ml of 14M HCl is required to achieve a pH of 9.00 in the solution.
Remember to always double-check your calculations and pay attention to significant figures when reporting your final answer.