A weak monoprotic acid has molar mass 180 g/mol. When 1.00 g of this acid is dissolved in enough water to obtain a 300 mL solution, the pH of the resulting solution is found to be 2.62. What is the value of Ka for this acid

The value of for the given acid is

Explanation:

To calculate the molarity of solution, we use the equation:

Initial mass of weak monoprotic acid = 1.00 g

Molar mass of weak monoprotic acid = 180 g/mol

Volume of solution = 300 mL

Putting values in above equation, we get:

To calculate the hydrogen ion concentration for given pH of the solution, we use the equation:

We are given:

pH = 2.62

Putting values in above equation, we get:

The chemical equation for the dissociation of weak monoprotic acid (HA) follows:

Initial:              0.0185

At eqllm:        0.0185-x     x     x

Evaluating the value of 'x'

So, equilibrium concentration of HA = (0.0185 - 0.0024) = 0.0161 M

Equilibrium concentration of = x = 0.0024 M

The expression of for above equation follows:

Putting values in above equation, we get:

Hence, the value of for the given acid is

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

the answer for sodium is 0.186

the answer for rubidium is 0.251