Question

The temperature for each solution is carried out at approximately 297 K where Kw=1.00×10−14.

0.25 g of hydrogen chloride (HCl) is dissolved in water to make 4.5 L  of solution. What is the pH of the resulting hydrochloric acid solution?

0.55 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 6.5 L of solution. What is the pH of this solution?

Answer 1

#

molar mass of HCl = 36.46094 gm/mole then 0.25 gm of HCl = 0.25 / 36.46094 = 0.0068567 mole

molarity = no. of mole / volume of solution in liter

molarity of HCl = 0.0068567 / 4.5 = 0.0015237 M

HCl is strong acid dissociate completly therefore [HCl] = [H⁺] = 0.0015237 M

pH = -log[H⁺] = -log(0.0015237) = 2.82

#

molar mass of NaOH = 39.997 gm/mole then 0.55 gm of NaOH = 0.55 / 39.997 = 0.01375 mole

molarity = no. of mole / volume of solution in liter

molarity of NaOH = 0.01375 / 6.5 = 0.0021155 M

NaOH is strong base dissociate completly therefore [NaOH] = [OH^-] = 0.0021155 M

pOH = -log[OH^-] = -log(0.0021155) = 2.67

pH = 14 - pOH = 14 - 2.67 = 11.33