Answer:
⇒ 3.312 E-13 mol Ni(OH)2 are soluble in 1 L of solution (NaOH)
Explanation:
S S 2S + 0.0218
∴ pKsp Ni(OH)2 = 15.8 = -Log Ksp
⇒ Ksp = 1.585 E-16 = [ Ni2+ ] * [ OH- ]²
∴ pH = 12.34
⇒ pOH = 14- 12.34 = 1.66 = - Log [ OH- ]
⇒ [ OH- ] = 0.0218 M
⇒ Ksp = 1.585 E-16 = S * ( 2S + 0.0218 )²
if we compare the concentration ( 0.0218 ) with the Ksp ( 1.585 E-16 ), then we can neglect the solubility as adding
⇒ 1.585 E-16 = S * ( 0.0218 )²
⇒ 1.585 E-16 = 4.786 E-4 * S
⇒ S = 3.312 E-13 mol/L
∴ soluble moles Ni(OH)2 = 3.312 E-13 mol/L * 1 L Sln = 3.312 E-13 moles
⇒ % S = ( 3.312 E-13 / 0.0218 ) * 100 = 1.513 E-11 %.....we can accept the previous assumption.
Explanation:
The solution of the lactic acd and sodium lactate is referred to as a buffer solution.
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. In this case, the weak acid is the lactic acid and the conjugate base is the sodium lactate.
Buffer solutions are generally known to resist change in pH values.
When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:
HA + OH- → A- + H2O.
Since the added OH- is consumed by this reaction, the pH will change only slightly.
The NaOH reacts with the weak acid present in the buffer sollution.
Density= mass/volume. volume= 12 x 6 x 10. 415/720=.57g/cm^3
First, we write our reaction and then we balance it:
AgBr + 2Na2S2O3 → Na3Ag(S203)2 + NaBr
To solve this, we need the molar masses of AgBr and NaBr (Please, use the periodic table to do this)
The molar mass of AgBr = 188 g/mol
The molar mass of NaBr = 103 g/mol
Procedure:
188 g AgBr --------------------- 103 g NaBr
41.3 g AgBr --------------------- X
X = 22.6 g
Answer: 22.6 g NaBr can be produced
