Answer:
0.500 M
Explanation:
The balanced equation for the neutralization reaction is as follows
NaOH + HCl —> NaCl + H2O
Molar ratio of NaOH to HCl is 1:1
Number of moles of NaOH reacted is equal to the number of HCl moles
We can use the following equation
c1v1 = c2v2
Where c1 is concentration and v1 is volume of HCl
c2 is concentration and v2 is volume of NaOH
Substituting the values
1.00 M x 10.0 mL = c2 x 20.0 mL
c2 = 0.500 M
Concentration of NaOH is 0.500 M
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Answer:
V₂ = 1518L
Explanation:
-Using combined gas law:

<em>Where P is pressure, V is volume and T is absolute temperature of 1, initial state and 2, final state of the gas.</em>
<em />
<em>Initial states:</em>
P₁ = 743 torr
V₁ = 975L
T₁ = 30°C + 273.15K = 303.15K
P₂ = 375 mmHg = torr
V₂ = Our incognite
T₂ = -35°C + 273.15K = 238.15K
Replacing:

<h3>V₂ = 1518L</h3>
Answer : The pH of buffer is 9.06.
Explanation : Given,

Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:


Therefore, the pH of buffer is 9.06.