The reaction continues but for our purposes we only care about the latter half of that equilibrium. In the question above, rather than carbonic acid, it is dealing with dissociation of phosphorous acid. The more Na the molecule has, the more hydrogen's are released. In the case of this reaction, the negative hco3- would with Na+ in solution making it neutral NaHCO3. This is analogous to carbonic acid deprotonating to become bicarbonate (h2co3 h+ + hco3-). The formula is actually just a simple dissociation/deprotonation of a diprotoc acid. This, in turn, decreases the pH of the solution.Ĭould you explain how you got the reaction formula? This increases the concentration of reactants, which will push the reaction right (forming more H+). Altering the ratio to favor the monosodium/disodium phosphate to favor the monosodium species is a fancy way of saying increasing the monosodium concentration. The general reaction is NaH2PO3 Na2HPO3 + H+ The more the reaction favors the dissociation of the monosodium phosphate, the more acidic the solution. The ratio does not change in the HH equation.Īs for why C is correct, look no further than our good friend LeChatlier. Adding strong base, diluting with water, or adding a different salt will NOT lower the pHįor MCAT purposes, diluting a solution does NOTHING to the pH because you are diluting A- and the HA the same amount. At low speed the force of deceleration will be small, so the rate at which velocity drops will be slow (shallow slope).Īlter the ratio of monosodium/disodium phosphate added to favor the monosodium speciesīecause in order to lower the pH of a buffer, the proportion of acidic buffer component must be increased. At high speed the force of deceleration will also be high, so the velocity will drop quickly (steep slope).
Slope of v, t graph = a, so in the answer choices the slope cannot stay the same because a is changing not constant, and it is decreasing overall since decelerationįirst to be clear, the passage states "Force decreases linearly with Speed" BUT the axes on the graphs are Velocity vs Time.Ĭhoice A would be if you had a CONSTANT decelerating forceĬhoice D shows the expected result of a LINEARLY DECREASING decelerating force. time graph, then the slope of the graph must decrease in timeį=ma, m will be constant, since force is decreasing linearly, acceleration is decreasing. Finally, because the decelerating acceleration is the rate of change of velocity, or the slope of the velocity vs. Because the decelerating force is directly proportional to the deceleration acceleration, it follows that the decelerating acceleration itself declines linearly with speed. Coz the speed of the car decreases in time such that the decelerating force declines linearly with speed, according to the passage.