Question

Chem. 18 wa t H Summary Name LONG 11/20/19 to 1. What is the pH of pure water toe 25 and 100C oc pth log (12 log 115wo 7.47 250hL 100 2. IHI 4.3, what is the pow at Cat C and at 100C OH43 PP24 Ans. 75,7.6.11 100°C pelas 53 -613 As 2 - 3. What is the pH of 0.01M H O.1MHNO, 10M HCL 1M HCIO: 0.0040M HCI? 4. What is the pH of 0.015M NOH; 0.25MKOH 5.0x10' M HCI: 3.0x10M HCIO 0.0040M HCI? 5. What is the pH of 50 ml 0.020M HCI? 0.05 * 0020 - 10 (1.1) 3 Ans. (12.6) 6. What is the pH of 100 ml 0.040M NOH? 0,1x0,04 410 port 1.44 pts 14-14..n.6. 7. What is the pH of 50 ml 0.010M HCI plus 50 ml of water? - 14- 14-16 05 0.01) - 14-3,3 = 10.7 c) pitt= 8. What is the pH of 0.40M weak acid with K. -1.0x10? Ans. 9. What is the pH of 50 ml of a 0.10M weak acid (K. -1.0x105) plus 50ml water? 10. What is the pH of 50 ml 0.030M weak base plus 100 ml water? Ko=1.0x10 Ans. . (10) 11. What is the pH of 50 ml 0.010M NaCl ? (Ans. 7.0 - salt of strong acid/strong base)

Answer 1

Solution, 8.1. What is the pH of Pure water at o'c, 25°C and 100°c. Ans Pure water is it self a very weak Electrolyte and ionised according to the following equation- H90 H+ + OH - The ionisation constant, ki is givenby. Kiz Cat. Con Since, wenter is found to be poonly ionised and it is concentral Can be combineed with the ionisation constant to give a new Constant, known as ionic Product of water kw! Henel, kw - Cit. Cu20 Cy, k - et . Сон- In Pure water Cat = COH.. KW = C4 = Col a r Cht = V kw . The Ko value following the temperature. I'c ko out 0c 0.114 x 10-14 porta CH+ = V kw = Jolly K 10-14 = 3.376810-8 .pH = -log Cu+ = 7:472 o 25°c 1:00x10-14 Cat = $100 x 10-14 = 10-7 pH = -log 44+ = 7. To'c 51.3x10-14 Cht= √51.3810-14 = 7.163610-7 pH = -log CH + = 6.14. 8.2. If ph is 4.3 What is the port at 0°c, at 25°C and at 100c? ³ we knew ph+ polt = 14 polt=9.71 NOTE: We Calculate pot at 25°c & 100°c, the Extra data one needed.

at is the PH of 0.01 (m) Hel; o.1 (M) HNO3, 10(m) Hel. 1 (m) Heloy and 0.0040 (m) Hel? Hene, Hel, Hm3, Heloy on one 13, Heloy on one strong coontorend monopnotic strong chords acid and aqueous solution they are full - solution they are ionized fullely D) 0.01(M) Hel: pH og Ht] = -log (0.01) = 2 001() Hyph Haz; pH = -log Cht] =- - Log (6.1) = 1. ma) 20cm) Hel; pH = -log (H+) = -log(10) = -4. N) 1 (m) Helly ; i; pH = -log [HT] = -log (1) = 0. pH. 47 v) 0.004(M) Hel; pH = -log [HT] = -log (0.004) = 2.4. 8.4. What is the pH of oat (A). Het ont in the motoren 0.015(M) NAH, 0.25 (m) kot 510-3 (m) Hel! 3x10-4 (M) Meloy 0.004 (m) Ha? we know ; pH + poH = 14. 0.015 (m) NOWOH; pOH = -log con ) = -log(0.015) =1824 .pH = 14-poh = 12176- i) 0.25 (m) KOH; pola-og (OH) = -log(0.25) = 0.602 .PH= 14-0.602 = 13.398. m) 5 x 10-3 (m) Hei; pH = -log (M+] = -og (5X10-3) = 2:3. iv) 3x10-4 (m) Heloy; pH = -log (147) -- log (3x10^4) = 3.523. V 0.004 (m) Hei; plta-log (14+) = -log co.coy) = 2.4 8.5. What is the pH of 50 mi 0.020(m) Melo As the pH is the measure of Concentration of a H+ ions and as concentration is an intensive property so pH is also an intensive Property.... . pH = -log(H+] = -log(0.02) - 1.7. 8.6. What is the pH of loomi 0.04(M) raut? Some Explanation of the question no.o. POH = -log [OH-] = -log (0.04) = 1.4. - P - 14 - P6 H = 16 - | : 4 = 26.

8.7. What is the pH of 50m 0.01 (m) Helt some of water? * 50m 0.01 (m) Hel solution Contarin 90 1x 50 = 5x10-4 molto 1000 Therefore, some of water added then the total volume = (50 +52) = 100 mi. Here, 100 mi solution Contarin 5*10-4 mol of her 1000 gr = 5610-4 x 1080 (m) of Hel 10 - 5x10-3 (M) of Het .pH = -log(Ht] = -log (5X10-3) = 2.3. 8.8. What is the pH of 0.4cm) weak deid with Ka = 1.08 10-50 the formula of pH of the week acid is. pH - - pm - log e. , але, po - - la ko -- {19 (1:0X10 -5 ) - į log (0.4) 215 +0.2 = 27. 8.9. what is the pH of 50 ml of a roon 0.10(m) w (ka = 10 x 10-5) +50 ml water. 50m weak acid solution added 50 m weiter, then the concentration of the weak aid = 0.1 = 0.05 (m). pH = 1 pka - 1/2 loge. - log (1.0x10-5) - { log 0.05 = 2.5 + 0.65 3.15. 8.10. What is the pH of 50m 0.03 (m) weak bage + 10o me of watero ko = 1.0x10-6. → 50 me of 0.03 cm) weak bage Contorn = 0.0 x 50 = 1.5x10-3 mal Again, 100 m of water added then the toten volume (50 + 100) = 150 m.

Bobil Solution contain 1.5810-3 mol weak base. 1000 v = 1.5*10-3 150X1000 (m) P = 0.01 (m) weak base . The formula of pot of the weak base- POH = 1 PKG - I loge where, pub= -logit. --{log (1-02 10–6) - 447 0:01 = 3+1=4. ..PH= 14-pol = 14-4 = 10. soll. What is the pH of 50 ml 0.010 (m) mael? Vael is a salt which obtain- strong acid + strong base. Here, The mad solution is the ovatnal solution. we knew, for the Nutral Solution pH=7. Here, pH of 50 m 0.010 (m) Nad is equart to 7. Nael -o-