How many grams of pure naoh must be used to prepare 10.0 l of a solution that has a ph of 13? ?

We must use 39.997 grams of NaOH to prepare 10.0 L of a solution that has a pH of 13.

To find the mass of NaOH, we need to calculate the number of moles.  

From the pH, we have:

The pOH is given by

Now, we can find the number of moles with the volume of solution

Finally, we can calculate the mass

Therefore, we need to use 39.997 g of NaOH.

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I hope it helps you!

40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.

Step 1. Calculate the pOH of the solution

pOH = 14.00 – pH = 14.00 -13 = 1

Step 2. Calculate the concentration of NaOH

[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L

Step 3. Calculate the moles of NaOH

Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH

Step 4. Calculate the mass of NaOH

Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.