There are several differences between<span> a </span>physical and chemical change<span> in matter or substances. A </span>physical change<span> in a substance doesn't </span>change<span> what the substance is. In a </span>chemical change<span> where there is a </span>chemical<span> reaction, a new substance is formed and energy is either given off or absorbed.</span>
I am first assuming that you want the molarity of the acid. Use the formula

which is the titration formula. M is molarity (a is of acid and b is of base) and V is volume in mL (a is of acid and b is of base). Plugging in gives us

. Solving gives us
Coarse and fine adjustment The coarse adjustment knob should only be used with the lowest power objective lens. Once it is in focus, you will only need to use the fine focus. Using the coarse focus with higher lenses may result in crashing the lens into the slide.
Answer:
a) 129.14 g/mol
b) 8.87
Explanation:
Given that:
mass of the ionic compound [NaA] = 18.08 g
Volume of water = 116.0 mL = 0.116 L
Let the mole of the acid HCl = 0.140 M
Volume of the acid = 500.0 mL = 0.500 L
pH = 4.63
= 1.00 L
Equation for the reaction can be represented as:

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L
= 0.140 mol
Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium
Thus, the molar mass of the sample = 
= 129.14 g/mol
b) since pH = pKa
Then pKa of HA = 4.63
Ka = ![10^{-4.63]](https://tex.z-dn.net/?f=10%5E%7B-4.63%5D)
= 
![[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}](https://tex.z-dn.net/?f=%5BA%5E-%5Dequ%20%3D%20%5Cfrac%7B0.140M%2A1.00L%7D%7B1.00L%2B0.116L%7D)

= 0.1255 M
of HA = 


+
+ 
Initial 0.1255 0 0
Change - x + x + x
Equilibrium 0.1255 - x x x
![K_b = \frac{[HA][OH^-]}{[A^-]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BHA%5D%5BOH%5E-%5D%7D%7B%5BA%5E-%5D%7D)
![4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}](https://tex.z-dn.net/?f=4.35%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.1255-x%5D%7D)
As
is very small, (o.1255 - x) = 0.1255

[OH⁻] = 
But pOH = - log [OH⁻]
= - log [
]
= 5.13
pH = 14.00 = 5.13
pH = 8.87