Materials that are eroded from one location are always ?

Which shows the formula for an Organic acid

tresset_1 [31]

Answer:
The formula of Organic acid is as follow,

R-COOH

Explanation:
The class of organic acids is called Carboxylic Acids. In above general structure, R is alkyl group and can vary. While -COOH is the functional group.
Carboxylic Acids has the tendency to loose protons and their pKa value depends upon the alkyl group. For example the pKa value of Acetic acid (R = -CH₃) is 4.7. The driving force for this acidity is the stability of carboxylate (conjugate base) due resonance. i.e

RCOOH ⇄ RCOO⁻ + H⁺
Where;
RCOO⁻ = Carboxylate Ion (Conjugate base)

3 0

3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$