Calculate the new concentration of NaOH if 35 mL of 2.0M NaOH is added to 45 mL of water.
If I don't use C1V1=C2V2, then how do I solve it? I'm confuused.
you are diluting it 80/35 =2.29 times. (one part original, 1.29 parts water).
So its new concentration : 2/(2.29) M=0.875M
A more conventional method. In the stsock solution, you had .035*2 moles of NaOH. You have that same amount in the final dilution.
Molarity=.035*2/.080=same as above.