Answers

2016-01-16T23:49:52+05:30
Best Answer:  HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) 

a) n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41×10ˉ³ mol 

n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13×10ˉ² mol 

NaOH is in excess, so amount remaining (greatest number of moles): 

1.13×10ˉ² - 8.41×10ˉ³ = 2.89×10ˉ³ mol 

Total volume = 0.0390 + 0.0290 = 0.068 L 

[OHˉ] = 2.89×10ˉ³ mol / 0.068 L = 4.25×10ˉ² M 

pOH = -log[OHˉ] = -log(4.25×10ˉ²) = 1.37 

pH = 14 - pOH = 14 - 1.37 = 12.6 

b) n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41×10ˉ³ mol 

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41x10ˉ³ mol 

HCl is in excess; amount remaining: 

8.41×10ˉ³ - 7.41×10ˉ³ = 1.0×10ˉ³ mol 

Total volume = 0.0290 + 0.0190 = 0.0480 L 

[HCl] = 1.0×10ˉ³ mol / 0.0480 L = 2.08×10ˉ² M = [H⁺] 

pH = -log[H⁺] = -log(2.08×10ˉ²) = 1.68
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2016-01-18T00:16:16+05:30
[OH-] in NaOH=10^-10 M We know [H+]×[OH-]=10^-14 [H+]=10^-14/10^-10=10^-4M pH=-log[H+] = -log10^-4 =-(-4log10)=-(-4×1)=4=pH
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