Question

please show work

1a) & 1b)

Question 1-a) What is the pH of a solution made by mixing 30.00 mL OF U. TUM HI WIL 40.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. A) 0.85 B) 1.85 C) 12.15 D) 13.15 *Answer: **Calculation: Question 1-b) What volume of 5.00 * 10-3 M HNO3 is needed to titrate 20.00 mL of 5.00 * 10-3 M Ca(OH)2 to the equivalence point? A) 2.50 mL B) 10.0 mL C) 20.0 mL D) 40.0 mL E) 180. mL *Answer: **Calculation: Page 1

Answer 1

The complete step wise calculation is given as below:-

- volume af acid = 30.0 mL = 0.030 L since 1L=1000mL concentration of acid = 0,10m. Number of moles of acid = (0.030L) *Colo meel 2 moles af acid = 0.03 moles - O volume af kol=40.0 m2=0.040L Concentration af kol = 0.1om Number of moles of kon = (0,000L) x (0.10 mol/L) moles afkoH = 0.04 moles of kon. Remaining moles of kor= (0.04 0.03 )= 0.01 moles Total volume = (0,03 +0,04L=0.07L

So, concentration af kol= (01ol moles 0.07L = 0.143 m So [ono = 0.143 m 3 pole - log [OHO = -log (0,143) = +0.845 sinee, pH = 14- pol = 14–0.845 = 13.155 or pn= 13:15 So, the correct option will be ① 13. 15.

6 It is given that, concentratio aftNo3 (61) = 5.00x103m volume af HNO3 (VI) =2 concentration of Ca (OH)2 = 5.000103 m = ca volume of Ca (OH) 2 = 2000 mL =V2 Then cixvi= Cq X V2 or v = C2 x V - 5000103 m × 2000mL 5.00 x 10m or Vi= 20.0 mL) since, I Ca(OH)2 gives a moles of one Then volume af Ca (OHq require = 20 = 1000m So, the correct option will be ® 10.0 mL.