Question

Calculate [OH ^-] and pH for each of the following solutions.



(a) 0.0088 M RbOH

[OH-] =  M

pH =

(b) 0.02 g of LiOH in 490.0 mL of solution

[OH -] =  M

pH =

(c) 89.8 mL of 0.00602 M Ca(OH)₂ diluted to 600 mL

[OH -] =  M

pH =

(d) A solution formed by mixing 51.0 mL of 0.000290 M Ca(OH)₂ with 83.0 mL of 6.8 x 10^-3 M RbOH

[OH -] =  M

pH =

Answer 1

@ Pooth is strong Bak v. CR bot] = lott] =o0offm pot = -log tolt-] = -log (0.0088) = 2.056 & pH = 4-pott a M-2.056 = 111.97 0.029 augo L 23.95 ging 0.00170 m TOH-] = pettle = log (OH = -log (0.00170) = 9.77 pH = 1a-pott - 14-2.77 ppt 1.2 & 11. 2 - 11. 2 male of Caalt) & mav. = 0.00620 mo/x 89.80123 0.0055676ndle

geteel udame - Gout 8 9.8= 689. f me = 0.689 1 After dilation - 0.0055677 mal NOW CACH), Canc 0.6898 L 0.0087 m = Caett, has exoho lang 2x 0.0080 Tro.0161m port = - log 16.0161)= 1.79, pH=14-179 Mole q CaCH)2 = 0.000 290 md 2x 51x153 = 1479x10 role role of RbelH} = 0.84 123 orel, s 3x1632 = 0.000sbury mole tofal mell of Tot ] = (2x 1.479x10+ 0.0005044 ) ale = 0.000 59398 mole fetel ledare - 51+ 8 3 * 134 m - 0.134 L Tot) = 0.00 59398 m2.00443m potte-loglolt-}-log (0.00 (143) = 2.35, pH= 14 -2.35=111.67 0 113UL