The NaOH will be used What titrant to titrate the 0. 02 m hcl phenol red solution.
Acid-base titrations may be the most typical titrations, although there are numerous more forms as well. Take a look at this illustration where sodium hydroxide is used to titrate a sample of hydrochloric acid (HCl) (NaOH). The titrant (NaOH), which is added gradually throughout the duration of the titration, has been added to the unknown solution.
Titrants are solutions with known concentrations that are added to solutions whose concentrations must be determined. The solution for whom the concentration needs to be determined is known as a titrant as well as analyte.
Therefore, the NaOH will be used as a titrant to titrate the 0. 02 m hcl phenol red solution.
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Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
<u>Answer:</u> The pH of resulting solution is 8.7
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of TRIS acid solution = 0.1 M
Volume of solution = 50 mL
Putting values in above equation, we get:
Molarity of TRIS base solution = 0.2 M
Volume of solution = 60 mL
Putting values in above equation, we get:
Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)
- To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7BTRIS%20base%7D%5D%7D%7B%5B%5Ctext%7BTRIS%20acid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of TRIS acid = 8.3
![[\text{TRIS acid}]=\frac{0.005}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20acid%7D%5D%3D%5Cfrac%7B0.005%7D%7B0.11%7D)
![[\text{TRIS base}]=\frac{0.012}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20base%7D%5D%3D%5Cfrac%7B0.012%7D%7B0.11%7D)
pH = ?
Putting values in above equation, we get:
Hence, the pH of resulting solution is 8.7
Answer:
Explanation:
use the equation
moles = mass/mr
=19.9/79.5
=0.250moles of CuO
then do the same for
H = 2.02/1
=2.02
so CuO is the limiting reagent because there is less amount of it.
Hope this helps :)
