Question

> Click Submit to complete this assessment. Question 5 of 5 Question 5 20 points Save Answer You add 7.50 mL of 0.0500 M NaOH to 25.00 mL of pure water, and to thig mixture you then add 1.50 ml of 0.200 M HCL What will be the pH of the resulting solution? 2.08 266 11,34 11.92 07.00 Question 5 of 5

Answer 1

No.of moles of NaOH = (7.5 mL) x (0.05 M) = 0.375 m.moles

No.of moles of HCl = (1.50 mL) x (0.200 M) = 0.300 m.moles

Hence, No.of moles of OH^- left unreacted = (0.375 - 0.300)m.moles = 0.075 m.moles

Total volume added = 7.5 mL + 1.5 mL + 25 mL = 34 mL

Hence, Concentration of OH^- = [OH^-] = (0.075 m.moles) / (34 mL) = 2.206 x 10^-3

Therefore, pH of the solution is given by:

pH = 14 - pOH

pH = 14 + log [OH^-]

pH = 14 + log (2.206 x 10^-3)

pH = 14 + (-2.656)

pH = 11.34 ----------- (answer)

• So the correct option is option (#3) : 11.34