Answer:
For disodium hydrogen phosphate:
5.32g Na2HPO4
For sodium dihydrogen phosphate:
7.65g Na2HPO4
Explanation:
First, you have to put all the data from the problem that you going to use:
-NaH2PO4 (weak acid)
-Na2HPO4 (a weak base)
-Volume = 1L
-Buffer pH = 7.00
-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M
What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:
H3PO4 = H2PO4 + H+ pKa1 = 2.14
H2PO4 = HPO4 + H+ pKa2 = 6.86
HPO4 = PO4 + H+ pKa3 = 12.4
So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:
pKa2= 6.86
Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:
pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]
The solution of the problem is attached to this answer.
Answer:
is a substance which donates an H^+ or a proton
There are two kinds of mixtures
a) homogeneous : the boundary of the two components is not physically distinct
b) heterogeneous:the boundary of the two components is physically distinct
the following separation techniques are common for mixtures
1) filtration: if the two components are forming heterogeneous mixture we can separate them by filtration.
2) boiling: if boiling point of one of the components is less than other
3) magnetic separation: if one of the component is magnetic
4)sieve method: for solid components with difference in size of particles
5) hand picking
Thus the correct match will be as shown in the figure
Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
Answer:
The boiling point is the temperature at which the vapor pressure equals the pressure of gas.
The normal boiling point is the temperature at which the vapor pressure equals one atmosphere
Explanation:
