The formula for the monoprotic acid is taken as HA, reaction with base is as follows;
HA + NaOH ---> NaA + H₂O
Stoichiometry of acid to base is 1:1
At the neutralisation point, number of HA moles = number of base moles
Number of NaOH moles reacted = 0.100M / 1000 mL /L x 30.0 mL = 0.003 mol
Therefore number of HA moles reacted = 0.003 mol
the mass of acid 0.384 g
Therefore molar mass - 0.384 g/ 0.003 mol = 128 g/mol
<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
<u>Explanation:</u>
The chemical equation for the dissociation of butanoic acid follows:

The expression of
for above equation follows:
![K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BCH_3CH_2CH_2COO%5E-%5D%5BH%5E%2B%5D%7D%7B%5BCH_3CH_2CH_2COOH%5D%7D)
We are given:
![[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COOH%5D%3D0.888M%5C%5CK_a%3D1.54%5Ctimes%2010%5E%7B-5%7D)
![[CH_3CH_2CH_2COO^-]=[H^+]](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COO%5E-%5D%3D%5BH%5E%2B%5D)
Putting values in above expression, we get:
![1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}](https://tex.z-dn.net/?f=1.54%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH%5E%2B%5D%5E2%7D%7B0.888%7D)
![[H^+]=-0.0037,0.0037](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D-0.0037%2C0.0037)
Neglecting the negative value because concentration cannot be negative
To calculate the volume of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is butanoic acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
C. The design would not be cost-efficient; the plane would require more fuel due to increased thrust from higher drag.
Answer:
False
Explanation:
The attractive force that holds atoms or ions together is called chemical bond.
Ba(OH)2 is an basic solution. It has more OH- ions than H+ ions. pOH should be calculated to find out its pH
The reaction is
Ba(OH)2 ⇒ Ba2+ (aq) + 2 OH-(aq)
One mole barium hydroxide releases 2 moles hydroxide ions.
Use that ratio to calculate molarity (M) of OH- ions [OH-]. The ratio is 1:2.
0.10 M Ba(OH)2 release 2*0.10 M= 0.02 M OH- ions
[OH-]= 0.02
pOH= - log [OH-] = - log 0.02 = 1.7
Thats not the answer! We found pOH of the solution before titration.
pH and pOH relationship is shown by formula of pH+pOH= 14
pH= 14-pOH
pH= 14-1.7= 12.3