Explanation:
According to the Henderson-Hasselbalch equation, the relation between pH and 
is as follows.
pH = 
where, pH = 7.4 and = 7.21
As here, we can use the nearest to the desired pH.
So, 7.4 = 7.21 + 
0.19 =
= 1.55
1 mM phosphate buffer means ![[HPO_{4}]](https://tex.z-dn.net/?f=%5BHPO_%7B4%7D%5D)
+ ![[H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
Therefore, the two equations will be as follows.
= 1.55 ............. (1)
+ = 1 mM ........... (2)
Now, putting the value of from equation (1) into equation (2) as follows.
1.55![[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D%20%2B%20%5Btex%5D%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
2.55 = 1 mM
= 0.392 mM
Putting the value of in equation (1) we get the following.
0.392 mM + = 1 mM
= (1 - 0.392) mM
= 0.608 mM
Thus, we can conclude that concentration of the acid must be 0.608 mM.
For the first one the answer is B. and the second one is D.
Answer: 1.24 × 10^25
Explanation:
× 
Using our knowledge in unit conversions, we know the mole units cancel each other out and all there's left is the atom unit. From here we can multiply the fractions and eventually we end with the number 124.0532 × 10^23
According to the scientific notation rules, the number to the left of the decimal cannot exceed 10 so we have to move the decimal to the left two spaces. With this change, we also have to change the exponent of the 10. Because we moved the decimal point two spaces to the left, that means we have 10^25.
26g --- 1 mol
56g --- X
X= 56/26 = 2,154 mol
959 ml = 959cm³ = 0,959dm³
C = n/V
C = 2,154/0,959
C = 2,246 mol/dm³
