Question

Calculate [OH-] and pH for each of the following solutions. (a) 0.0061 M KOH [OH-] = C M pH = (b) 0.0225 g of KOH in 540.0 mL of solution [OH-] = pH = (C) 53.0 mL of 0.00788 M Sr(OH), diluted to 700 mL [OH ") = C M pH=C (d) A solution formed by mixing 44.0 mL of 0.000590 M Sr(OH)2 with 25.0 mL of 3.2 x 10-3 M KOH [OH ) = C M pH =

Calculate [OH ^-] and pH for each of the following solutions.
(a) 0.0061 M KOH

[OH-] = ? M pH=?

(b) 0.0225 g of KOH in 540.0 mL of solution

[OH-] = ? M pH=?

(c) 53.0 mL of 0.00788 M Sr(OH)₂ diluted to 700 mL

[OH-] = ? M pH=?

(d) A solution formed by mixing 44.0 mL of 0.000590 M Sr(OH)₂ with 25.0 mL of 3.2 x 10^-3 M KOH

[OH-] = ? M pH=?

Answer 1

а) о- доб| мкон Кон — кә - бі— - Сон) — о. oo6 | Арөн = — wg tон) — — Log to oo 67) - 2-21 .: pH = 14 – Но — 124 — 2-2) = 11:19 .

S40 (b) Molar mom of kot = 56.1056 gmots Q-org kон 4 кон —— 4-01 624 аз хоч m() 4 kbp. • 540 ml ѕh tortum -o10 29 473 х 10 Ч mi+ kola 1800 mL son contron 4.01029 4873x10-4 x 1000 mol of kot e 7.43 x 10-4 mol of kolt. A cont-4 кон – 143х10 ч то/ — 1:43 Хоч ту KOH kt ton- : онг) = 143х согу м. pOH =- wgton) 2 - (-93 X102 — 3-13 pH = 1 - 2 он = 14-3-13 — 10-87 ) VS = v 5,— у 53 ml xo-001 м = to mi X, ул - 67 yil y 0 - 007 ft и то от 15. -БТ х 10-ум A LOH), ) = S. 99 х 0-9 м Зr (o 1), нү2+ + 0 — 5. 47х10 Чм S-ча хії 'm axs. Чах 10-9 м

[OH = 245.97810-4m. = 1-194 x103 m 04 Stop tou s pot = -log tot] = -log (1.1948103 - 2-92 pH = 14 -polt = 14-2.92 - 11.08 (d) " 44 MLX 0.000590m. TSCOH Lorcom (44+25) = 3.762318841X10-Y m Sr Coll) 2 + 204 (OH) = 293.762318841 X10-4 m = 7.52 4 6 3768 1 8 104 m. (hot) = 25mly 3.2x183 m. (25+44 m = 1.159.42029 X10-3 m KOH K+ + OH- in TOH) = 1.159 420 29x103 m. Total concentration of ot- LOH) = 47.524637 684 X104)+(1.15942042X6037 – 1.91x 103 m e port = – log tot) = wg (1.91x103) – 2.72 pH = 14-polt - 14-2.72 = 11-28 w