A. how many grams of KOH should you add to 1L of water to make a 2 molarity solution

b. what is the pH of this solution?

a. 2 moles x 56.11 g/mole= 112.22 g KOH per liter
b. [OH]- = 2 moles/L
pH= 14 - (-log[OH-]
pH=13.7

A. 2(molarity) = x(moles)/1(liters)
multiply 1 on both sides to get x by it self, x = 2(moles)
1 mole of KOH = 39(K) + 16(O) +1(H) = 56 grams. 56 grams per mole
2 moles = 56 x 2 = 112 grams.

B, KOH ----> K + OH
molarity of KOH = 2, soo molarity of OH = 2.
-log(2) = pOH = -0.30. doesn't make sense because you can't find the negative log and it would make the pH = 14.30 when it only goes 14, but thats what i got.

are you sure it said 1L and not 1mL?

You need molarity -> C=2
Volume -> 1L
number of moles = x
C=n/v
2=x/1
x=2moles of KOH
MW(KOH)=59.10grams/mol
m=n*MW=59.1*2=118.2grams

KOH ---> K+ + OH-
Strong base disociates completly
n(OH-)=2moles
C(OH-)=2
POH=-LOG[2]=0.3
PH=14-POH=14-0.3=13.7

molar mass KOH = 56.1g/mol
2M solution = 56.1*2 = 112.2g KOH in 1 litre of solution = 2.0M solution

KOH is a strong base
[OH-] = 2.0M
pOH = -log [OH-]
pOH = -log2.0
pOH = - 0.30

pH = 14 - pOH
pH = 14.00-(-0.30)
pH = 14.00+0.30
pH = 14.30