Hello!
First, we need to determine the pKa of the base. It can be found applying the following equation:
Now, we can apply the
Henderson-Hasselbach's equation in the following way:
![pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BCH_3NH_2%5D%7D%7B%5BCH_3NH_3Cl%5D%7D%20%29%3D10%2C65%2Blog%28%20%5Cfrac%7B0%2C18M%7D%7B0%2C73M%7D%20%29%3D10%2C04)
So,
the pH of this buffer solution is 10,04Have a nice day!
Answer:
copper is the only material composed of a single pure substance
Explanation:
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Answer:
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Explanation:
Answer:
#1: 0.00144 mmolHCl/mg Sample
#2: 0.00155 mmolHCl/mg Sample
#3: 0.00153 mmolHCl/mg Sample
Explanation:
A antiacid (weak base) will react with the HCl thus:
Antiacid + HCl → Water + Salt.
In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.
Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:
Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl
Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl
Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl
The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.
Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:
#1: 2.178mmol / 1515mg
#2: 2.253mmol / 1452mg
#3: 2.219mmol / 1443mg
<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>
