Answer:- The hydroxide ion concentration of the solution is 
.
Solution:- The formula used to calculate pOH from hydroxide ion is:
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:
![[OH^-]=10^-^p^O^H](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5Ep%5EO%5EH)
pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:
![[OH^-]=10^-^5^.^7^1](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5E5%5E.%5E7%5E1)
![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
= 0.00000195 or
So, the hydroxide ion concentration of the solution is .
I believe this is a double replacement. As you can see none of the chemical symbols are missing, there just "replaced".
Answer:
55.9 g KCl.
Explanation:
Hello there!
In this case, according to the definition of molality for the 0.500-molar solution, we need to divide the moles of solute (potassium chloride) over the kilograms of solvent as shown below:
Thus, solving for the moles of solute, we obtain:
Since the density of water is 1 kg/L, we obtain the following moles:
Next, since the molar mass of KCl is 74.5513 g/mol, the mass would be:
Regards!
