Can someone please calculate the ph of the solutions below????????? ONLY 5 QUESTIONS?

You need to use:

ph = - log M or pOH = -log M ( use this to basics)

1. 0.01 M HCl

pH = - log 0.01 / pH = 2

2. 0.0010 M NaOH

pOH = - log M / pOH = 3 / use this relation: pOH + pH = 14 / so, 3 + pH = 14 / pH = 11

3. 0.050 M Ca(OH)2 ( this, you need to observe that rate of OH liberated, see: Ca(OH)2 ---> Ca^+2 + 2 OH- , so you have 2 OH- being liberated, so the concentration of OH- is gonna be 2 times the concentration of Ca(OH)2.

so, pOH = - log M / pOH = - log 0,1 / pOH = 1,1 / pOH + pH = 14 / 1,1 + pH = 14 / pH = 12,9

4. 0.030 M HBr

pH = - log 0,030 / pH = 1,52

5. 0.150 M KOH

pOH = -log M / pOH = 0,823 / pOH + pH = 14 / pH = 13, 17

Observation: when you are dealing with basics, you need to find the pOH fist, than you convert to pH.

Second: you need to observe the stoichiometry. I mean, you need to observe how many mols are being formed of the basics or acid.

Take care.

Paulo, from Brazil.

Why don't you do it?