Answer:
Final pH: 9.49.
Round to two decimal places as in the question: 9.5.
Explanation:
The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.
What's the pKb of base B?
Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.
.
.
.

.
What's the new salt-to-base ratio?
The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.
Initial:
;
.
After adding the HCl:
;
.
Assume that the volume is still 0.5 L:
.
.
What's will be the pH of the solution?
Apply the Henderson-Hasselbalch equation again:
![\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BpOH%7D%20%3D%20%5Ctext%7BpK%7D_b%20%2B%20%5Clog%7B%5Cfrac%7B%5B%5Ctext%7BSalt%7D%5D%7D%7B%5B%5Ctext%7BBase%7D%5D%7D%7D%20%3D%204.64613%20%2B%20%5Clog%7B%5Cfrac%7B0.760%7D%7B1.04%7D%7D%20%3D%204.50991)
.
The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.
Answer:
The anion of the compound is b CH3COO-
Explanation:
The compound cupric acetate Cu(CH3COO)2 undergo dissociation to form cupric ion Cu2+ and Acetate anion(CH3COO-)
Cu(CH3COO)2⇒ Cu2+ + CH3COO-
From the above equation it can be stated that the anion of Cu(CH3COO)2 is CH3COO-.
Answer:
0.095M
Explanation:
HClO4 + NaOH = NaClO4 + H2O
Concentration of acid CA= 0.170M
Concentration of base CB= ???
Volume of base VB= 28.5ml
Volume of acid VA= 16.0ml
Number of moles of acid nA= 1
Number of moles of base nB= 1
From
CAVA/CBVB= nA/nB
CB= CAVAnB/VBnA
CB= 0.170×16.0×1/28.5×1
CB= 0.095M
<span>La temperatura de congelación del agua es 0 grados </span>centígrados.