What is the pH of 500 mL of water after the addition of 1 mL of 2M NaOH?

pH = 11.60

Explanation:

When we add a base, we are increasing [OH⁻], so the pH will be totally basic.

pH of water = 7

Basic pH > 7

We are adding 2 M . 1 mL = 2 mmoles

2 mmoles of OH⁻ are contained in 501 mL of total volume.

[OH⁻] = 2 mmol / 501 mL =  3.99×10⁻³ M

- log 3.99×10⁻³ M = 2.39 → pOH

pH = 14 - pOH → 11.61

Water equilibrium

2H₂O  ⇄  H₃O⁺   +  OH⁻          Kw = 1×10⁻¹⁴

NaOH  →  Na⁺   +  OH⁻

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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