Question 5.
The no. of moles of HCl = 2 mol/L * (250/1000) L = 0.5 mol
The no. of moles of NaOH = 0.5 mol/L * (150/1000) L = 0.075 mol
The resulting no. of moles of HCl after getting neutralized with NaOH = 0.5 - 0.075 = 0.425 mol
The total volume of the resulting solution = 250 mL + 150 mL = 400 mL = 0.4 L
Therefore, the [H+] of the resulting solution = 0.425 mol/0.4 L = 1.0625 M
Hence, pH = -Log[H+] = -Log(1.0625) = -0.026
Consider the mat foundation in the figure below. L-12 m B-25000 k D 1.5 m, x,-2 m, x 3 m, X3-4 m. The clay is...
125 mL of 0.010 M HI. 16.45 Calculate [OH] and pH for (a) 1.5 x 103MSr(OH) 2, ב- (b) 2.250 g of LIOH in 250.0 mL of solution, (c) 1.00 mL of 0.175 M NAOH diluted to 2.00 L, (d) a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 102 M Ca (OH)2.