2020-08-09T02:49:09-04:00
A solution is prepared from 20mL of 0.10M HNO3 and 30mL of 0.05M H2SO4. If the H2SO4 completely ionises, the pH of the solution would be:
1
2020-08-17T13:34:55-0400
n 1 ( H + ) = c ( H N O 3 ) V 1 = 0.10 M ∗ 0.02 L = 0.002 m o l n_1(H^+) = c(HNO_3)V_1 = 0.10M*0.02L = 0.002mol n 1 ( H + ) = c ( H N O 3 ) V 1 = 0.10 M ∗ 0.02 L = 0.002 m o l
n 2 ( H + ) = 2 c ( H 2 S O 4 ) V 2 = 2 ∗ 0.05 M ∗ 0.03 L = 0.003 m o l n_2(H^+)=2c(H_2SO_4)V_2 = 2*0.05M*0.03L = 0.003mol n 2 ( H + ) = 2 c ( H 2 S O 4 ) V 2 = 2 ∗ 0.05 M ∗ 0.03 L = 0.003 m o l
n t o t a l = 0.002 m o l + 0.003 m o l = 0.005 m o l n_{total} = 0.002mol+0.003mol = 0.005mol n t o t a l = 0.002 m o l + 0.003 m o l = 0.005 m o l
V t o t a l = 0.02 L + 0.03 L = 0.05 L V_{total} = 0.02L+0.03L = 0.05L V t o t a l = 0.02 L + 0.03 L = 0.05 L
c ( H + ) = n / V = 0.005 m o l / 0.05 L = 0.10 M c(H^+) = n/V = 0.005mol/0.05L = 0.10M c ( H + ) = n / V = 0.005 m o l /0.05 L = 0.10 M
p H = − l o g c ( H + ) = + 1 pH = -logc(H^+) = +1 p H = − l o g c ( H + ) = + 1 Answer : A +1
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