Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:
acts as a weak acid and 
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:
Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:
Taking antilog:
On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is required to make the buffer is 0.138M.
Answer:
6
Explanation:
You will see H6 and the H stands for helium and the 6 is how many of that atom is there
Answer:
Ar < Cl - < S2-
Explanation:
All the species written above are isoelectronic. This means that they all possess the same number of electrons. All the species above possess 18 electrons, the noble gas electron configuration.
However, for isoelectronic species, the greater the atomic number of the specie, the smaller it is. This is because, greater atomic number implies that their are more protons in the nucleus exerting a greater attractive force on the electrons thereby making the specie smaller in size due to high electrostatic attraction.
<span>There
are a number of ways to express concentration of a solution. This includes
molarity. Molarity is expressed as the number of moles of solute per volume of
the solution. We calculate the mass of the solute by first determining the number of moles needed. And by using the molar mass, we can convert it to units of mass.
Moles </span>(nh4)3po4 = 0.250 L (0.150 M) = 0.0375 moles (nh4)3po4
Mass = 0.0375 mol (nh4)3po4 (149.0867 g / mol) = 5.59 g (nh4)3po4
According to Henderson–Hasselbalch Equation,
pH = pKa + log [Lactate] / [Lactic Acid]
As,
Ka of Lactic Acid = 1.38 × 10⁻⁴
pKa = -log Ka
pKa = -log 1.38 × 10⁻⁴
pKa = 3.86
So,
pH = 3.86 + log [0.10] / [0.13]
pH = 4.74 + log 0.769
pH = 4.74 - 0.11
pH = 4.63
