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![@munuat iem CH2 (eJ -4.-ー ト 2 じNah(01) 6. 346 PH 6,276 Ans pka= 6.396 0.6. 2 2</ 외엔6J@oH (O1] 는 o 6224](https://i.imgur.com/f2uvtDR.png)
Given H_2CO_3 rightarrow 1 mol volume of solution = 1L concentration [H_2CO_3] = 1/1 = mol/volume of solution = 1mol NaHCO_3 rightarrow 1mol [NaHCo_3] = 1/1 = 1mol/L pH = p^ka + log_10 [[NaHCO_3]/[H_2CO_3]] p^ka = -log [k_a] implies - log (4.2 times 10^-7) = 7 - log (4.2) = 6.376 p_H = 6.376 + log_10^1 p_H = 6.376 Ans [H_2CO_3] = 0.100m volume = 1 liter k_a = 4.2 times 10^-7 p^ka = 6.376 p_H = p^ka + log_10 [[NaHCO_3]/[H_2CO_3]] 7 - 6.376 = log_10 [[NaHCO_3]/0.100] 0.6224 = log_10 [[NaHCO_3]/0.100] 10^0.6244 times 0.100 = [NaHCO_3] [NaHCO_3] = 0.419 \r\n mole of NaHCO_3 = [NaHCO_3] times volume = 0.419mol Ans mole concentration of [NH_3] = 0.01M mole concentration of [NH_4Cl] = 0.05M p^k = -log_10 [k_b] = -log_10 (1.8 times 10^-5) = 5 - log 1.8 = 5 - 0.255 = 4.744 pH = 14 - [pk_b + log_10 [NH_4Cl]/[NH_3]] = 14 - [4.744 + log_10 [0.05/0.01] =
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